RD Sharma Class 12 Exercise 30.4 Probability Solutions Maths-Download PDF Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 04:02 PM IST

RD Sharma is viewed as the best material for CBSE students. They are utilized all around the nation and are very useful for exam planning. The Class 12 RD Sharma chapter 30 exercise 30.4 solution is one of the examples for it. They are liked over NCERT because of their prospectus inclusion and detail of replies. The RD Sharma class 12th exercise 30.4 Probability can be utilized for self-practice and assessment of imprints.

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RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise
Probability Exercise: 30.4
RD Sharma Chapter-wise Solutions

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise

Probability Exercise: 30.4

Probability Exercise 30.4 Question 1(i)

Answer: A and B are independent events
Given: A coin is tossed thrice and all the eight outcomes are assumed equally likely.
A= the first throw results in head, B= the last throw results in tail
Hint: Check, $P\left ( A\: \cap\: B \right )=P\left ( A \right )P\left ( B \right )$
Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\: \cap\: B \right )=P\left ( A \right )P\left ( B \right )$
Let,
$\begin{aligned} &S=\left\{\begin{array}{l} (\mathrm{HHH}),(\mathrm{HHT}),(\mathrm{HTH}),(\mathrm{HTT}) \\ (\mathrm{THH}),(\mathrm{THT}),(\mathrm{TTH}),(\mathrm{TT}) \end{array}\right\} \\ \end{aligned}$
$\begin{aligned} &\mathrm{P}(\mathrm{A})=\frac{4}{8}=\frac{1}{2} \\ &\mathrm{P}(\mathrm{B})=\frac{4}{8}=\frac{1}{2} \\ &\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{8}=\frac{1}{4} \\ &\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}) \end{aligned}$
Thus, A and B are independent events.

Probaility Exercise 30.4 Question 1(ii)

Answer: A and B are not independent events
Given: A coin is tossed thrice and all the eight outcomes are assumed equally likely.
A= the number of heads is odd, B= the number of tails is odd
Hint: Check, $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$ or not
Solution: : As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Let,
$\begin{aligned} &S=\left\{\begin{array}{l} (\mathrm{HHH}),(\mathrm{HHT}),(\mathrm{HTH}),(\mathrm{HTT}) \\ (\mathrm{THH}),(\mathrm{THT}),(\mathrm{TTH}),(\mathrm{TT}) \end{array}\right\} \\ \end{aligned}$
$\begin{aligned} &\mathrm{P}(\mathrm{A})=\frac{4}{8}=\frac{1}{2} \\ &\mathrm{P}(\mathrm{B})=\frac{4}{8}=\frac{1}{2} \\ &\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{0}{8}=0 \\ \end{aligned}$
Here $\begin{aligned} &\mathrm{P}(\mathrm{A} \cap \mathrm{B})\neq \mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}) \end{aligned}$

Thus, A and B are not independent events.

Probability Exercise 30.4 Question 1(iii)

Answer: A and B are not independent events
Given: A coin is tossed thrice and all the eight outcomes are assumed equally likely.
Hint: Check, whether $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$ or not
Solution: : As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
$\begin{aligned} &S=\left\{\begin{array}{l} (\mathrm{HHH}),(\mathrm{HHT}),(\mathrm{HTH}),(\mathrm{HTT}) \\ (\mathrm{THH}),(\mathrm{THT}),(\mathrm{TTH}),(\mathrm{TT}) \end{array}\right\} \\ \end{aligned}$
$\begin{aligned} &\mathrm{P}(\mathrm{A})=\frac{3}{8} \\ &\mathrm{P}(\mathrm{B})=\frac{4}{8}=\frac{1}{2} \\ \end{aligned}$
Now, $\begin{aligned} &\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{8}=\frac{1}{4} \\ \end{aligned}$
$\begin{aligned} &\mathrm{P}(\mathrm{A} \cap \mathrm{B})\neq \mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}) \end{aligned}$

A and B are not independent events.

Probability Exercise 30.4 Question 2

Answer: A and B are independent events
Given: Prove that in throwing a pair of dice, the occurrence of the number 4 on the first die is independent of the occurrence 5 on the second die.
Hint: $Probaility =\frac{No\: of\: outcomes}{Total\: Outcomes}$
Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Here, Total number of events = 36
A = The event of 4 on first die
B = Event of 5 on second die
$\begin{aligned} &P(A)=\frac{6}{36}=\frac{1}{6} \ldots \ldots .\{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\} \\ &P(B)=\frac{6}{36}=\frac{1}{6} \ldots \ldots \ldots \ldots . .\{(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)\} \\ &P(A \cap B)=\frac{1}{36} \ldots \ldots \ldots \ldots \ldots . .\{(4,5)\} \end{aligned}$
$P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Hence,A and B are independent events

Probability Exercise 30.4 Question 3(i)

Answer: A and B are not independent events
Given: A card is drawn from a pack of 52 cards so that each card is equally likely to be selected.
A= the card is a king or queen, B= the card drawn is a queen or jack.
Hint: $Probaility=\frac{No\: of\: outcomes}{Total\: Outcomes}$
Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Let A = Event of King or Queen
$P\left ( A \right )=\frac{8}{2}=\frac{2}{13}$ ….{No. of king and Queen are 4+4=8 and total outcomes=52}
B = Event of Queen or Jack
$P\left ( B \right )=\frac{8}{56}=\frac{2}{13}$ ……{ No. of Jack and Queen are 4+4=8 and total outcomes=52}
$P\left ( A\cap B \right )=P\left ( Queen \right )$
$=\frac{4}{12}=\frac{1}{13}$ …{No. of queen in a pack of cards =4}
$P\left ( A\cap B \right )\neq P\left ( A \right )P\left ( B \right )$
Thus, A and B are not independent events.

Probability Exercise 30.4 Question 3(ii)

Answer: A and B are independent events
Given: A card is drawn from a pack of 52 cards so that each card is equally likely to be selected.
A= the card is black, B= the card drawn is a king
Hint: To find events are independent $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Let, A =Event of getting a black card
$P\left ( A \right )=\frac{Number\: of\: Black\: Cards}{Total\: Number\: of\: Cards}$
$=\frac{26}{52}.....(Total \: Number\: of\: black\: cards=26)$
$=\frac{1}{2}$
B = Event of getting an king
$P\left ( B \right )=\frac{Number\: of\: Black\: Cards}{Total\: Number\: of\: Cards}$
$=\frac{4}{52}$
$=\frac{1}{13}$
$P\left ( A\cap B \right )=\frac{2}{52}.....\left \{ N.\: of\: black\: kings=2 \right \}$
$P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Thus, A and B are independent events.

Probaility Exercise 30.4 Question 3(iii)

Answer: A and B are independent events
Given: A card is drawn from a pack of 52 cards so that each card is equally likely to be selected.
A= the card is a spade, B= the card drawn is an ace
Hint: Check $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
A = No. of cards is spade
$P\left ( A \right )\frac{13}{52}=\frac{1}{4}...{Total \: no.\: of \: spades=13}$
B = No. of cards is ace
$P\left ( B \right )=\frac{4}{52}=\frac{1}{13} \cdot \cdot \cdot \left \{ Total no. of ace cards=4 \right \}$
$P\left ( A\cap B \right )=\frac{1}{52}\left \{ {Total no. of ace of spade=1} \right \}$
$P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Thus, A and B are independent events.

Probability Exercise 30.4 Question 4(i)

Answer: A and B are independent events
Given: A coin is tossed three times. A = first toss is head, B= second toss is head, C=exactly two heads are tossed in a row. Check the independence of A and B
Hint: Check whether A and B are independent or not,
Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
$\begin{aligned} &\mathrm{S}=\left\{\begin{array}{l} (\mathrm{HHH}),(\mathrm{HHT}),(\mathrm{HTH}),(\mathrm{HTT}), \\ (\mathrm{THH}),(\mathrm{THT}),(\mathrm{TTH}),(\mathrm{TT}) \end{array}\right\} \\ \end{aligned}$
$\begin{aligned} &P(A)=\frac{4}{8}=\frac{1}{2} \\ &P(B)=\frac{4}{8}=\frac{1}{2} \\ &P(A \cap B)=\frac{2}{8}=\frac{1}{4}=P(A) P(B) \end{aligned}$
A and B are independent events

Probaility Exercise 30.4 Question 4(ii)

Answer: B and C are not independent events
Given: A coin is tossed three times. A = first toss is head, B= second toss is head, C=exactly two heads are tossed in a row. Check the independence of B and C
Hint: Check $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Let,
$\begin{aligned} &\mathrm{S}=\left\{\begin{array}{l} (\mathrm{HHH}),(\mathrm{HHT}),(\mathrm{HTH}),(\mathrm{HTT}), \\ (\mathrm{THH}),(\mathrm{THT}),(\mathrm{TTH}),(\mathrm{TT}) \end{array}\right\} \\ \end{aligned}$
$\begin{aligned} &P(A)=\frac{2}{8}=\frac{1}{4} \\ &P(B)=\frac{4}{8}=\frac{1}{2} \\ &P(B \cap C)=\frac{2}{8}=\frac{1}{4}\neq P(A) P(B) \end{aligned}$
Thus, B and C are not independent events.

Probability Exercise 30.4 Question 4(iii)

Answer: A and C are independent events
Given: A coin is tossed three times. A = first toss is head, B= second toss is head, C=exactly two heads are tossed in a row. Check the independence of C and A
Hint: Check $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
$\begin{aligned} &\mathrm{S}=\left\{\begin{array}{l} (\mathrm{HHH}),(\mathrm{HHT}),(\mathrm{HTH}),(\mathrm{HTT}), \\ (\mathrm{THH}),(\mathrm{THT}),(\mathrm{TTH}),(\mathrm{TTT}) \end{array}\right\} \\ \end{aligned}$
$\begin{aligned} &P(A)=\frac{2}{8}=\frac{1}{4} \\ &P(B)=\frac{4}{8}=\frac{1}{2} \\ &P(C \cap A)=\frac{1}{8}=P(C) P(A) \end{aligned}$

A and C are independent events

Probaility Exercise 30.4 Question 5

Answer: A and B are independent events
Hint: Use formula $P\left ( A\cap B \right )=P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )$

Given:
$P\left ( A \right )=\frac{1}{4}$
$P\left ( B \right )=\frac{1}{3}$
$P\left ( A\cap B \right )=\frac{1}{2}$

Solution:

As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$

$P\left ( A\cap B \right )=P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )$

$=\frac{1}{4}+\frac{1}{3}-\frac{1}{2}$

$=\frac{3+4-6}{12}$

$P\left ( A\cap B \right )=\frac{1}{12}=\frac{1}{4}\times P\left ( A \right )P\left ( B \right )$

Thus, A and B are independent events.

Probability Exercise 30.4 Question 6(i)

Answer: $0.18$
Hint: Use $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Given,
$P\left ( A \right )=0.3$
$P\left ( B \right )=0.6$
Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$

$P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
$=\left ( 0.3 \right )\left ( 0.6 \right )$
$=0.18$

Probability Exercise 30.4 Question 6(ii)

Answer: $0.12$
Hint: Use, $P\left ( A\cap \bar{B} \right )=P\left ( A \right )P\left ( \bar{B} \right )$
Given,
$P\left ( A \right )=0.3$
$P\left ( B \right )=0.6$
Solution:
$P\left ( A\cap \bar{B} \right )=P\left ( A \right )P\left ( \bar{B} \right )$
$=\left ( 0.3 \right )\left [ 1-P\left ( B \right ) \right ]$
$=\left ( 0.3 \right )\left [ 1-0.6 \right ]$
$=\left ( 0.3 \right )\times \left ( 0.4 \right )$
$0.12$

Probability Exercise 30.4 Question 6(iii)

Answer: $0.42$
Hint: $P\left ( \bar{A}\cap B \right )=P\left ( \bar{A} \right )P\left ( B \right )$
Here A and B are independent events
Given:
$P\left ( A \right )=0.3$
$P\left ( B \right )=0.6$
Solution:
$P\left ( \bar{A}\cap B \right )=P\left ( \bar{A} \right )P\left ( B \right )$
$\begin{aligned} &=(P(B))[1-P(A)] \\ &=(0.6)[1-0.3] \\ &=(0.6) \times(0.7) \\ &=0.42 \end{aligned}$

Probability Exercise 30.4 Question 6(iv)

Answer: $0.28$
Hint:Use
$\begin{gathered} \mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\overline{\mathrm{B}}) \\ \end{gathered}$
Given:
$\begin{gathered} \mathrm{P}(\mathrm{A})=0.3 \\ \mathrm{P}(\mathrm{B})=0.6 \\ \end{gathered}$
Solution:
$\begin{gathered} \mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\overline{\mathrm{B}}) \\ \end{gathered}$
$\begin{gathered} =[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})] \\ \end{gathered}$
$\begin{gathered} =[1-0.3][1-0.6] \\ \end{gathered}$
$\begin{gathered} =(0.7) \times(0.4) \\ \end{gathered}$
$=0.28$

Probaility Exercise 30.4 Question 6(v)

Answer: $0.72$
Hint: Using,
$P\left ( A\cap B \right )=P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )$
Given,
$P\left ( A \right )=0.3$
$P\left ( B \right )=0.6$

Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
$P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )=0.3*0.6=0.18$
$\begin{aligned} P(A \cup B) &=P(A)+P(B)-P(A \cap B) \\ &=0.3+0.6-0.18 \\ &=0.9-0.18 \\ &=0.72 \end{aligned}$

Probaility Exercise 30.4 Question 6(vi)

Answer: $0.3$
Hint: Using $P\left ( \frac{A}{B} \right )=\frac{P\left ( A\cap B \right )}{P\left ( B \right )}$
Given,
$P\left ( A \right )=0.3$
$P\left ( B \right )=0.6$
Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
$\begin{aligned} P\left(\frac{A}{B}\right) &=\frac{P(A \cap B)}{P(B)} \\ &=\frac{P(A) P(B)}{P(B)} \\ &=0.3 \end{aligned}$

Probability Exercise 30.4 Question 6(vii)

Answer: $0.6$
Hint: Use $P\left ( \frac{B}{A} \right )=\frac{P\left ( B\cap A \right )}{P\left ( A \right )}$
Given,
$P\left ( A \right )=0.3$
$P\left ( B \right )=0.6$
Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
$\begin{aligned} P\left(\frac{B}{A}\right) &=\frac{P(B \cap A)}{P(A)} \\ &=\frac{P(A) P(B)}{P(A)} \\ &=0.6 \end{aligned}$

Probability Exercise 30.4 Question 7

Answer: $0.77$
Hint: $P\left ( A \right )+P\left ( \bar{A} \right )=1$
Given,
$P\left ( \bar{B}\right ) =0.65$
$P\left ( A\cup B \right )=0.85$
Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
$\begin{aligned} P(B) &=1-P(\bar{B}) \\ &=1-0.65 \\ &=0.35 \\ \end{aligned}$
Now,
$\begin{aligned} P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ \end{aligned}$
$\begin{aligned} 0.85 &=P(A)[1-P(B)]+P(B) \\ 0.85 &=P(A) P(\bar{B})+0.35 \\ 0.85 &-0.35=P(A) \times 0.65 \\ \frac{0.50}{0.65} &=P(A) \\ P(A) &=\frac{10}{13} \\ &=0.77 \end{aligned}$

Probability Exercise 30.4 Question 8

Answer: $y=\frac{4}{5}$ or $\frac{1}{6}$
Hint: Form events as an equation
Given:
$P\left ( \bar{A}\cap B \right )=\frac{2}{15}$
$P\left ( A\cap \bar{B} \right )=\frac{1}{16}$
$P\left ( A \right )$ & $P\left ( B \right )$ are independent events
Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Let,
$\begin{aligned} &\mathrm{P}(\mathrm{A})=\mathrm{x} \\ &\mathrm{P}(\mathrm{B})=\mathrm{y} \\ &\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})=\frac{2}{15} \\ &\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\mathrm{B})=\frac{2}{15} \\ &(1-\mathrm{x}) \mathrm{y}=\frac{2}{15} \end{aligned}$ ............(1)
Also
$\begin{aligned} &P(A \cap \bar{B})=\frac{1}{6} \\ &P(A) P(\bar{B})=\frac{1}{6} \\ &x(1-y)=\frac{1}{6} \end{aligned}$ ........(2)

Subtracting (1) from (2)

$\begin{aligned} &y-x y=\frac{2}{15} \\ &x-x y=\frac{1}{6} \\ &y-x=\frac{2}{15}-\frac{1}{6} \\ &y-x=\frac{-3}{90} \\ &y+\frac{1}{30}=x \end{aligned}$

Substitute value of x in (2)

$\begin{aligned} &\left(y+\frac{1}{30}\right)(1-y)=\frac{1}{6} \\ &30 y^{2}-29 y+4=0 \end{aligned}$

Solve using quadratic formula

$\begin{aligned} &a=30, b=-29, c=4 \\ &\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a} \\ &\quad=\frac{-(-29)+\sqrt{841-480}}{60} \\ \end{aligned}$

$\begin{aligned} &\alpha=\frac{29+19}{60} \\ &\alpha=\frac{48}{60} \quad \\ &\alpha=\frac{4}{5} \end{aligned}$

Similarly $\begin{aligned} \beta &=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a} \\ \end{aligned}$

$\begin{aligned} &=\frac{29-19}{60} \\ &=\frac{10}{60} \\ &=\frac{1}{6} \\ &y=\frac{4}{5} \text { or } \frac{1}{6} \end{aligned}$

Probability Exercise 30.4 Question 9

Answer: $\begin{aligned} &P(B)=\frac{1}{2} \text { then } P(A)=\frac{1}{3} \\ \end{aligned}$
$\begin{aligned} &P(B)=\frac{1}{3} \text { then } P(A)=\frac{1}{2} \end{aligned}$

Hint: Form events as a equation to solve

Given, A and B are independent events

$P\left ( A\cap B \right )=\frac{1}{6}$

$P\left ( \bar{A}\cap \bar{B} \right )=\frac{1}{3}$

Solution:

As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$

Now,

$\begin{aligned} &P(A \cap B)=P(A) P(B) \\ &\frac{1}{6}=P(A) P(B) \\ &P(A)=\frac{1}{6(P(B))} \end{aligned}$ ...................(1)

Now, for $P(\bar{A} \cap \bar{A})=P(\bar{A}) P(\bar{B}) \\$

$\begin{aligned} &\frac{1}{3}=[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})] \\ &\frac{1}{3}=\left[1-\frac{1}{6 \mathrm{P}(\mathrm{B})}\right][1-\mathrm{P}(\mathrm{B})] \\ &\frac{1}{3}=\left[1-\frac{1}{6 \mathrm{x}}\right][1-\mathrm{x}] \end{aligned}$

Where, $P\left ( B \right )=x$

$\begin{aligned} &\frac{1}{3}=\left[\frac{6 x-1}{6 x}\right][1-x] \\ &2 x=6 x-6 x^{2}-1+x \\ &2 x-6 x+6 x^{2}+1-x=0 \\ \end{aligned}$

$\begin{aligned} &6 x^{2}-5 x+1=0 \\ &6 x^{2}-3 x-2 x+1=0 \\ &3 x(2 x-1)-1(2 x-1)=0 \\ &(2 x-1)(3 x-1)=0 \\ \end{aligned}$

Probability Exercise 30.4 Question 10

Answer: $P\left ( B \right )=0.5$
Hint: Use $P ( A \cap B)=P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )$
Given,
$P\left ( A \right )=0.2$
$P ( A \cap B)=0.60$
Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P ( A \cap B)=P\left ( A \right )P\left ( B \right )$
Let, $P\left ( B \right )=x$
$\begin{aligned} &P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ &P(A \cup B)=P(A)+x-P(A) P(B) \\ &P(A \cup B)=P(A)[1-x]+x \\ \end{aligned}$
$\begin{aligned} &0.60=0.2-0.2 x+x \\ &0.60-0.2=0.8 x \\ &x=\frac{0.40}{0.80} \\ &x=0.5 \\ &P(B)=0.5 \end{aligned}$

Probability Exercise 30.4 Question 11

Answer: $\frac{1}{4}$
Given: A die is tossed twice. Find the probability of getting a number greater than 3 on each toss.
Hint: $Probability=\frac{No\: of\: outcomes}{Total \: Outcomes}$
Solution:
Let, S= Total no. of dice in two throws
$\mathrm{S}=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}$
$n\left ( S\right )=36$
E=Getting a number greater than 3 on each toss
$\begin{aligned} &\mathrm{E}=\left\{\begin{array}{l} (4,4),(4,5),(4,6),(5,4),(5,5) \\ ,(5,6),(6,4),(6,5),(6,6) \end{array}\right\} \\ &\mathrm{n}(\mathrm{E})=9 \\ &\begin{aligned} \mathrm{P}(\mathrm{E}) &=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})} \\ &=\frac{9}{36}=\frac{1}{4} \end{aligned} \end{aligned}$

Probability Exercise 30.4 Question 12

Answer: $\frac{2}{15}$
Hint: Use, $P\left ( \bar{A}\cap \bar{B} \right )=P\left ( \bar{A} \right )P\left ( \bar{B} \right )$
Given: A can solve the problem = $\frac{2}{3}$
B can solve the problem= $\frac{3}{5}$
Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
A can solve the problem = $\frac{2}{3}$
i.e., $P\left ( A \right )=\frac{2}{3}$
B can solve the problem= $\frac{2}{3}$
i.e., $P\left ( B \right )=\frac{3}{5}$
$\begin{aligned} &\mathrm{P}(\mathrm{A})+\mathrm{P}(\overline{\mathrm{A}})=1 \\ &\mathrm{P}(\overline{\mathrm{A}})=1-\frac{2}{3} \\ &=\frac{1}{3} \end{aligned}$
Similarly, $P\left ( \bar{B} \right )=1-P\left ( B \right )$
$=1-\frac{3}{5}$
$=\frac{2}{5}$
$P\left ( \bar{A}\cap \bar{B} \right )=P\left ( \bar{A} \right )P\left ( \bar{B} \right )$
$=\frac{1}{3}\times \frac{2}{5}$
$=\frac{2}{15}$

Probability Exercise 30.4 Question 13

Answer: $\frac{1}{3}$
Hint: Use $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Given:
A = No. 4, 5 or 6 on first toss
B = 1, 2, 3 or 4 on second toss
Solution:
$\mathrm{S}=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}$
$n\left ( S \right )=36$
$\begin{aligned} &P(A)=\frac{18}{36}=\frac{1}{2} \\ &P(B)=\frac{24}{36}=\frac{2}{3} \\ &\begin{aligned} P(A \cap B) &=P(A) P(B) \\ &=\frac{1}{2} \times \frac{2}{3} \\ &=\frac{1}{3} \end{aligned} \end{aligned}$

Probability Exercise 30.4 Question 14

Answer: $\frac{9}{25},\frac{4}{25},\frac{6}{25}$
Hint: $Probability=\frac{No\: of\: outcomes}{Total\: outcomes}$
Given, A Bag contains 3 red & 2 black balls. One ball is drawn at random. Its color is noted and then it is put back in the bag. A second drawn is made and the same procedure is repeated.
Solution:
Let 3 red balls be $R_{1},R_{2},$ & $R_{3}$ and 2 black balls be $B_{1}$ & $B_{2}$ .
Total Possibilities = $\begin{gathered} \left\{\left(R_{1}, R_{2}\right),\left(R_{1}, R_{3}\right),\left(R_{1}, B_{1}\right),\left(R_{1}, B_{2}\right),\left(B_{1}, B_{2}\right),\left(R_{2}, R_{1}\right) \right. \\ \end{gathered}$
$\begin{gathered} \left(R_{2}, R_{3}\right),\left(R_{2}, B_{1}\right),\left(R_{2}, B_{2}\right),\left(R_{3}, R_{1}\right),\left(R_{3}, R_{2}\right),\left(R_{3}, B_{1}\right),\left(R_{3}, B_{2}\right),\left(R_{1}, R_{1}\right),\left(R_{2}, R_{2}\right),\left(R_{3}, R_{3}\right) \\ \left.\left(B_{1}, B_{1}\right),\left(B_{2}, B_{2}\right),\left(B_{1}, R_{1}\right),\left(B_{1}, R_{2}\right),\left(B_{1}, R_{3}\right),\left(B_{2}, R_{1}\right),\left(B_{2}, R_{2}\right),\left(B_{2}, R_{3}\right),\left(B_{2}, B_{1}\right)\right\} \end{gathered}$
i. Probability of drawing two red balls $=\frac{9}{25}$
ii. Probability of drawing two black balls $=\frac{4}{25}$
iii. Probability of first red and second black $=\frac{6}{25}$

Probability Exercise 30.4 Question 15

Answer: $\frac{6}{2197}$
Hint: Use permutations for finding ${ }^{n} P_{r}=\frac{n !}{(n-r) !}$
Given:
Three cards are drawn with replacement from a well shuffled pack of cards. Find the probability that the cards drawn are king, queen and jack.
Solution:
Let,
A= Event of getting kings
$P\left ( A \right )=\frac{No.\: of\: cards\: of\: king}{Total\: no.of\: cards}=\frac{4}{52}$
Similarly, B= Event of getting queens
$P\left ( B \right )=\frac{No.\: of\: cards\: of\: queens }{Total\: no.of\: cards}=\frac{4}{52}$
C= Event of getting jacks
$P\left ( C \right )=\frac{No.\: of\: jack\: cards\: }{Total\: no.of\: cards}=\frac{4}{52}$
Probability that cards drawn are king, queen and jack $=\frac{4}{52}\times \frac{4}{52}\times \frac{4}{52}\times 3P_{2}$
(Permutations has been used because the order of king,queen and jack can be changed)
$\begin{aligned} &=\frac{4 \times 4 \times 4 \times 6}{52 \times 52 \times 52} \\ &=\frac{6}{2197} \end{aligned}$

Probability Exercise 30.4 Question 16

Answer: $0.8645$
Given: An article manufactured by a company consists of two parts X and Y. In the process of manufacture of the part X, 9 out of 100 parts may be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part Y.
Hint: $Probability=\frac{No\: of\: outcomes}{Total\: outcomes}$
Solution:
Let, A= Particle X is defective
B= Particle Y is defective
$P\left ( A \right )=\frac{9}{100}$
$P\left ( B \right )=\frac{5}{100}$
Required Probability $=P\left ( \bar{A}\cap \bar{B} \right )$
$\begin{aligned} &=\mathrm{P}(\overline{\mathrm{A}}) \times \mathrm{P}(\overline{\mathrm{B}}) \\ &=[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})] \\ &=\left[1-\frac{9}{100}\right]\left[1-\frac{5}{100}\right] \\ &=\frac{91}{100} \times \frac{95}{100} \\ &=0.8645 \end{aligned}$

Probability Exercise 30.4 Question 17

Answer: $\frac{3}{5}$
Hint: Use, $P\left ( A\cup B \right )=P(A)+P\left ( B \right )-P\left ( A\cap B \right )$
Given, Possibilities
$P\left ( A \right )=P$ (A hits target) $=\frac{1}{3}$
$P\left ( B \right )=P$ (B hits target) $=\frac{2}{5}$

Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
$P\left ( A \cup B \right )=P$ (target will be hit by either A or B)
$\begin{aligned} P(A \cup B) &=P(A)+P(B)-P(A \cap B) \\ P(A \cup B) &=P(A)+P(B)-P(A) P(B) \\ &=\frac{1}{3}+\frac{2}{5}-\frac{1}{3} \times \frac{2}{5} \\ &=\frac{5+6-2}{15} \\ &=\frac{9}{15} \\ &=\frac{3}{5} \end{aligned}$

Probability Exercise 30.4 Question 18

Answer: $0.6976$
Hint: $P\left ( A \right )+P\left ( \bar{A} \right )=1$
Given: An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively.
Solution:
Let,
P (Gun hits the plane) = 1 – (Gun does not hit the plane)
$\begin{aligned} &P(A)=1-P(\bar{A}) \\ &\begin{aligned} P(\bar{A}) &=(1-0.4)(1-0.3)(1-0.2)(1-0.1) \\ &=0.6 \times 0.7 \times 0.8 \times 0.9 \\ &=0.3024 \\ P(A) &=1-0.3024 \\ P(A) &=0.6976 \end{aligned} \end{aligned}$

Probability Exercise 30.4 Question 19

Answer: i. $\frac{52}{77}$ ii. $\frac{25}{77}$
Hint: $\begin{aligned} P(A \cup B) &=P(A)+P(B)-P(A \cap B) \\ \end{aligned}$
$\begin{aligned} &=P(A)+P(B)-P(A) P(B) \\ &=P(A)[1-P(B)]+P(B) \end{aligned}$
Given: The odds against event A are 5 to 2. The odds in flavor of event B are 6 to 5.
Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $\begin{aligned} P(A \cap B) &=P(A)P(B)\end{aligned}$
The odds against event A are 5 to 2.
$P(A)=\frac{\text { no.of outcome }}{\text { total outcome }}=\frac{2}{5+2}=\frac{2}{7}$

The odds in flavor of event B are 6 to 5

$P(B)=\frac{\text { no.of outcome }}{\text { total outcome }}=\frac{6}{6+5}=\frac{6}{11}$

Probability of atleast one of the event occurs $\Rightarrow$

$\begin{aligned} P(A \cup B) &=P(A)+P(B)-P(A \cap B) \\ &=\frac{2}{7}+\frac{6}{11}-\frac{2}{7} \times \frac{6}{11} \\ &=\frac{22+42-12}{77} \\ &=\frac{52}{77} \end{aligned}$

ii P(none of the event occurred) $=1-P\left ( A\cup B \right )$

$=1-\frac{52}{77}$

$=\frac{25}{77}$

Probability Exercise 30.4 Question 20

Answer: $\frac{7}{8}$
Hint: $P(A \cup B \cup C)=P(A)+P(B)+P(C)-$ $[P(A \cap B)+P(B \cap C)+P(C \cap A)]+P(A \cap B \cap C)$
Given: A die is thrown thrice. Find the probability of getting an odd number at least once.
Solution:
P (Getting an odd number in one throw) $=\frac{3}{6}=\frac{1}{2}$ …{there are 3 odd numbers in an throw of a dice}
Here, getting an odd number in three throws refers to 3 independent events
$P\left ( A \right )=P\left ( B \right )=P\left ( C \right )=\frac{1}{2}$
$P(A \cup B \cup C)=P(A)+P(B)+P(C)-$ $[P(A \cap B)+P(B \cap C)+P(C \cap A)]+P(A \cap B \cap C)$
$\begin{aligned} &=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}-\left[\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2}\right]+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \\ &=\frac{12-6+1}{8} \\ &=\frac{7}{8} \end{aligned}$

Probability Exercise 30.4 Question 21

Answer: i. $\frac{16}{81}$ ii. $\frac{20}{81}$ iii. $\frac{40}{81}$
Hint: $Probability=\frac{No.\: of\: outcomes}{Total\: outcomes}$
Given: Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.
Solution:
$\begin{aligned} &\text { Total balls }=10 \text { black }+8 \text { red }=18 b a l l s \\ \end{aligned}$
$\begin{aligned} &P(\text { first red ball })=\frac{8}{18} \\ \end{aligned}$
$\begin{aligned} &\left.P(\text { second red ball })=\frac{8}{18} \ldots \text { fas replacement is allowed }\right\} \\ \end{aligned}$
$\begin{aligned} &P(\text { first ball is black })=\frac{10}{18} \\ &P(\text { sec ond ball is black })=\frac{10}{18} \ldots\{\text { as replacement is allowed }\} \end{aligned}$
$i. \quad P( two red balls )=\frac{8}{18} \times \frac{8}{18}=\frac{16}{81}$
$ii. \quad \mathrm{P}( first \: ball\: is \: black \: and\: second \: is \: red )=\frac{10}{18} \times \frac{8}{18}=\frac{20}{81}$
$iii. \mathrm{P} (one of them is black and other is red) =\mathrm{P} (first ball is red \& second is black)$
$+\mathrm{P}( first ball is black \& second is red)$
$\begin{aligned} &=\frac{8}{18} \times \frac{10}{18}+\frac{10}{18} \times \frac{8}{18} \\ &=\frac{40}{81} \end{aligned}$

Probability Exercise 30.4 Question 22(i)

Answer: $\frac{16}{121}$
Hint: $Probability=\frac{No\: of\: outcomes}{Total\: outcomes}$
Given: An urn contains 4 red and 7 black balls. Two balls are drawn at random with replacement. Find the probability of getting 1) 2 red balls
Solution:
$Total\: balls = 4 red+7 black=11 balls$
$\begin{aligned} \mathrm{P}(2 \text { red balls }) &=\mathrm{P}(\text { first ball is red }) \times \mathrm{P}(\text { second ball is red }) \\ &\left.=\frac{4}{11} \times \frac{4}{11} \ldots \text { as replacement is allowed }\right\} \\ &=\frac{16}{121} \end{aligned}$

Probability Exercise 30.4 Question 22(ii)

Answer: $\frac{49}{121}$
Hint: $Probability=\frac{No\: of\: outcomes}{Total\: outcomes}$
Given: An urn contains 4 red and 7 black balls. Two balls are drawn at random with replacement. Find the probability of getting 1) 2 black balls
Solution:
$Total\: \: balls = 4 red+7 black=11 balls$
$\begin{aligned} P(2 \text { black balls }) &=P(\text { first is black }) \times P(\text { second is black }) \\ =& \frac{7}{11} \times \frac{7}{11} \ldots\{\text { as replacement is allowed }\} \\ \end{aligned}$
$\begin{aligned} &=\frac{49}{121} \end{aligned}$

Probability Exercise 30.4 Question 22(iii)

Answer: $\frac{56}{121}$
Hint: $Probability=\frac{No\: of\: outcomes}{Total\: outcomes}$
Given: An urn contains 4 red and 7 black balls. Two balls are drawn at random with replacement. Find the probability of getting 1) one red and one black ball
$Total\: \: balls = 4 red+7 black=11 balls$
$P\left ( One\: red\: \&\: One Black \right )$
$\begin{aligned} &=P(\text { first red and second black })+P(\text { first black and second } \mathrm{red}) \\ &=\frac{4}{11} \times \frac{7}{11}+\frac{4}{11} \times \frac{7}{11} \\ &=\frac{56}{121} \end{aligned}$

Probability Exercise 30.4 Question 23

Answer: $\frac{26}{49}$
Hint: $probability=\frac{No\: of\: outcomes}{Total\: outcomes}$
Given: The probabilities of two students A and B coming to the school in time are
$\frac{3}{7}$ & $\frac{5}{7}$
Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
$\begin{aligned} &P(A \text { coming in time })=\frac{3}{7} \\ &P(\text { A not coming in time })=1-\frac{3}{7}=\frac{4}{7} \\ &P(B \text { coming in time })=\frac{5}{7} \\ &P(B \text { not coming in time })=1-\frac{5}{7}=\frac{2}{7} \\ \end{aligned}$
$\begin{aligned} &P(\text { only one of } A \& B \text { comin } g \text { in time })=P(A) P(\bar{B})+P(\bar{A}) P(B) \\ \end{aligned}$
$\begin{aligned} &=\frac{3}{7} \times \frac{2}{7}+\frac{4}{7} \times \frac{5}{7} \\ &=\frac{26}{49} \end{aligned}$

Probability Exercise 30.4 Question 24

Answer: No pair is independent
Hint: $Probability=\frac{No\: of\: outcomes}{Total\: outcomes}$
Given: Two dice are thrown together and the total score is noted. The event E,F and G are “a total 4”, “a total of 9 or more” and “ a total divisible by 5”
Solution:
$S=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}$
$\begin{aligned} &n(S)=36 \\ &E=a \text { total of } 4 \\ &\quad=\{(1,3),(2,2),(3,1)\} \\ &\text { ie, } n(E)=3 \end{aligned}$

$F=a\: total\: of\: 9\: or\: more$

$=\left\{\begin{array}{l} (3,6),(4,5),(4,6),(5,4),(5,5) \\ (5,6),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}$

i.e,n(f)=10

$G=a\: total\: divisible\: by\: 5$

$=\left\{\begin{array}{l} (1,4),(2,3),(3,2),(4,1), \\ (4,6),(5,5),(6,4) \end{array}\right\}$

i.e,n $(G)=7$

Now,

$\begin{aligned} &P(E)=\frac{3}{36}=\frac{1}{12} \\ &P(F)=\frac{10}{36}=\frac{5}{18} \\ &P(G)=\frac{7}{36} \end{aligned}$

Also,

$\begin{aligned} &E \cap F=\varnothing, E \cap G=\varnothing \\ &F \cap G=\{(4,6),(5,5),(6,4)\} \\ &\text { ie, } n(F \cap G)=3 \\ \end{aligned}$

$\begin{aligned} &P(F) \times P(G)=\frac{5}{18} \times \frac{7}{36} \\ \end{aligned}$

$\begin{aligned} &=\frac{35}{648} \\ &\neq \frac{3}{36} \end{aligned}$

$i.e,P(F)\times P\left ( G \right )\neq P\left ( F\cap G \right )$

So, no pair is independent.

Probability Exercise 30.4 Question 25(i)

Answer: $P_{1}P_{2}=P\left ( A \& B Occur \right )$
Given: Let A and B be two independent events such that
$P(A)=p_{1} \& P(B)=p_{2}$
Hint: Use, $P\left ( A \right )\times P\left ( B \right )=P\left ( A\cap B \right )$
Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
As, $P_{1}P_{2}=P\left ( A \right )\times P\left ( B \right )$
And A and B are independent events
i.e $\begin{gathered} P(A) \times P(B)=P(A \cap B) \\ \end{gathered}$
$\begin{gathered} P(A \cap B)=P_{1} P_{2} \end{gathered}$
Hence, $P_{1}P_{2}=P\left ( A \& B Occur \right )$

Probability Exercise 30.4 Question 25(ii)

Answer: $\left ( 1-P_{1} \right )P_{2}=P$ (A does not occur ,but B Occurs)
Given: Let A and B be two independent events such that
$P\left ( A \right )=p_{1}\&P\left ( B \right )=p_{2}$
Hint: Use $P\left ( A \right )+P\left ( \bar{A} \right )=1$
Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
$\begin{aligned} &\left(1-P_{1}\right) P_{2}=[1-P(A)] \times P(B) \\ \end{aligned}$
$\begin{aligned} &=P(\bar{A}) \times P(B) \\ \end{aligned}$
And A and B are independent events
$\begin{aligned} &\text { ie, } \mathrm{P}(\overline{\mathrm{A}}) \times \mathrm{P}(\mathrm{B})=\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B}) \\ &\text { So, } \mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})=\left(1-\mathrm{P}_{1}\right) \mathrm{P}_{2} \end{aligned}$
Hence, $\left ( 1-P_{1} \right )P_{2}=P$ (A does not occur ,but B Occurs)

Probability Exercise 30.4 Question 25(iii)

Answer: $1-\left ( 1-P_{1} \right )\left ( 1-P_{2} \right )=P\left ( Atleast\: one\: of\: A\: and\: B\: occurs \right )$
Given: Let A and B be two independent events such that
$P\left ( A \right )=p_{1}\&P\left ( B \right )=p_{2}$
Hint: Use, $P\left ( \bar{A}\right )\times P\left ( \bar{B} \right )=P\left ( \bar{A}\cap \bar{B} \right )$
Solution:
As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$
Given, A and B are independent events
As, $\begin{aligned} 1-\left(1-P_{1}\right)\left(1-P_{2}\right) &=1-[1-P(A)][1-P(B)] \\ \end{aligned}$
$\begin{aligned} &=1-P(\bar{A}) \times P(\bar{B}) \end{aligned}$
And A and B are independent vectors
$\begin{aligned} &\mathrm{P}(\overline{\mathrm{A}}) \times \mathrm{P}(\overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) \\ &\begin{aligned} \Rightarrow 1-\left(1-\mathrm{P}_{1}\right)\left(1-\mathrm{P}_{2}\right)=& 1-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) \\ &=1-\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B}}) \\ &=\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \end{aligned} \\ \end{aligned}$
So, $\begin{aligned} &P(\mathrm{~A} \cup \mathrm{B})=1-\left(1-\mathrm{P}_{1}\right)\left(1-\mathrm{P}_{2}\right) \end{aligned}$
Hence, $1-\left ( 1-P_{1} \right )\left ( 1-P_{2} \right )=P\left ( Atleast\: one\: of\: A\: and\: B\: occurs \right )$

Probability Exercise 30.4 Question 25(iv)

Answer: $\mathrm{P}_{1}+\mathrm{P}_{2}-2 \mathrm{P}_{1} \mathrm{P}_{2}=\mathrm{P}( Exactly\: \: one \: \: of \mathrm{A} \& \mathrm{~B} occurs )$
Given: Let A and B be two independent events such that
$P(A)=p_{1} \& P(B)=p_{2}$
Hint: $P(A) \times P(B)=P(A \cap B)$
Solution: As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e. $P(A \cap B)=P(A) P(B)$
Given,
$\begin{aligned} &P_{1}+P_{2}-2 P_{1} P_{2}=\left(P_{1}-P_{1} P_{2}\right)+\left(P_{2}-P_{1} P_{2}\right) \\ \end{aligned}$
$\begin{aligned} &=[P(A)-P(A) \times P(B)]+[P(B)-P(A) \times P(B)] \end{aligned}$
And A and B are independent events
i.e $P(A) \times P(B)=P(A \cap B)$
$\begin{aligned} \Rightarrow \mathrm{P}_{1}+\mathrm{P}_{2}-2 \mathrm{P}_{1} \mathrm{P}_{2} &=[\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})]+[\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})] \\ &=\mathrm{P}(\text { only } \mathrm{A})+\mathrm{P}(\text { only } \mathrm{B}) \end{aligned}$
$So, \mathrm{P}( only \mathrm{A})+\mathrm{P}( only mathrm{B})=\mathrm{P}_{1}+\mathrm{P}_{2}-2 \mathrm{P}_{1} \mathrm{P}_{2}$
Hence, $\mathrm{P}_{1}+\mathrm{P}_{2}-2 \mathrm{P}_{1} \mathrm{P}_{2}=\mathrm{P}( Exactly\: \: one \: \: of \mathrm{A} \& \mathrm{~B} occurs )$

Probability Exercise 30.4 Question 26

Answer: $\frac{126}{1000}$
Given:
$Probability\: of\: finding\: a\: green\: signal\: on\: X=30%$
Hint: $Probability=\frac{No\: of\: outcomes}{Total\: outcomes}$
Solution:
Given,
$Probability\: of\: finding\: a\: green\: signal\: on\: X=\frac{30}{100}=\frac{3}{10}$
$Probability\: of\:not\: finding\: a\: green\: signal\: on\: X=1-\frac{30}{100}=\frac{7}{10}$
Now, Let $D_{1},D_{2},D_{3}$ be the three days for which the signal is green and $D{}'_{1},D{}'_{2},D{}'_{3}$ be the three days for not finding a green signal.
We have to find probability of finding green signal on consecutive two days out of three
So the event would be $\left ( D_{1},D_{2},D{}'_{3} \right )$ or $\left ( D{}'_{1},D_{2},D_{3} \right )$
So the probability of finding green signal on consecutive two days out of three
$\begin{aligned} &=\mathrm{P}\left(D_{1}, D_{2}, D_{3}^{\prime}\right)+P\left(D_{1}^{\prime}, D_{2}, D_{3}\right) \\ &=\frac{3}{10} \times \frac{3}{10} \times \frac{7}{10}+\frac{7}{10} \times \frac{3}{10} \times \frac{3}{10} \\ &=\frac{63}{1000}+\frac{63}{1000} \\ &=\frac{126}{1000} \end{aligned}$

The RD Sharma class 12 solution of Probability exercise 30.4, is utilized by thousands of students and educators for commonsense math information. The RD Sharma class 12th exercise 30.4 has 43 inquiries like Autonomous Occasions, Similarly possible Occasions, Comprehensive occasions, Association and crossing point, with substitution and without substitution based inquiries.

The benefits of practicing from the RD Sharma class 12th exercise 30.4 are referred to beneath: -

You can use RD Sharma class 12th exercise 30.4 material will assist students with improving comprehension of the chapter and score gold imprints in exams.
Career360 RD Sharma class 12 chapter 30 exercise 30.4 trusted in the scrutinizing material that different understudies have used. Teachers in like manner are particularly attached to the book concerning the subject of maths.
RD Sharma class 12th exercise 30.4 is given via Career360 free of cost for students to concentrate on problem-free.
RD Sharma class 12th exercise 30.4 solutions made by specialists and are refreshed to the most recent form of the book, which implies there would be no missing inquiries or additional inquiries.

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Frequently Asked Question (FAQs)

1. What is the best mathematics reference material available online for class 12 students?

The presence of Class 12 RD Sharma Chapter 30 Exercise 30.4 is a gift to the current batch students. They must utilize it in the right way to score good marks.

2. Where can I find all the answers for the questions given in the 4th exercise of chapter 30?

The RD Sharma Class 12 Chapter 30 Solution provides the solved sums for all the 43 questions in it.

3. Are there any updates made in the RD Sharma solution books?

Frequent updates take place according to the changes made in the mathematics textbook.

4. In which website is the latest collection of RD Sharma reference books available?

The students can find the recent collection of the RD Sharma reference materials at the Career360 website.

5. Should only the class 12 students use the RD Sharma solution books?

The RD Sharma solution books can be used by class 11, 12 students and even teachers, tutors who handle classes for them.