RD Sharma Solutions Class 12 Mathematics Chapter 30 FBQ

Edited By Satyajeet Kumar | Updated on Jan 25, 2022 05:33 PM IST

When a student enters Class 12, selecting the suitable material for preparation becomes crucial and scoring well academically. The Rd Sharma Class 12th Exercise FBQ solutions will assist students in developing better skills and efficiently preparing for exams. Students should go through chapter 30 thoroughly because Probability is one of the most important topics to learn in maths. Our subject matter experts have prepared solutions and even designed the chapter 30 problems so that you can have fun while solving them.

RD Sharma Class 12 Solutions Chapter30 FBQ Probability - Other Exercise

Probability Excercise:FBQ

Probability Exercise Fill in the Blanks Question, Question 1.

Answer:

Answer: $P(A)P(\overline{B})$
Given:
If A clip_image004

and B

are independent events then P(A∪B)=1-x clip_image008

where x=…

. -
Hint:
Using P(A)=1-P(A) ; P(A)=1-P(A) clip_image012

Explanation:
We are given
A and $\overline{B}$ an indepindent events
$\Rightarrow \overline{A}$ and B oure also independent
Since $\overline{A}$ and B are independent events.
$\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})=\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\mathrm{B})$
Now
$\begin{aligned} &\mathrm{P}(\overline{\mathrm{A}} \cup \mathrm{B})=\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B}) \\ &=(1-\mathrm{P}(\mathrm{A}))+(1-\mathrm{P}(\overline{\mathrm{B}}))-\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\mathrm{B}) \\ &=(1-\mathrm{P}(\mathrm{A}))+(1-\mathrm{P}(\overline{\mathrm{B}}))-\cdot(1-\mathrm{P}(\mathrm{A}))(1-\mathrm{P}(\overline{\mathrm{B}})) \end{aligned}$
$\begin{aligned} &=1-\mathrm{P}(\mathrm{A})+1-\mathrm{P}(\overline{\mathrm{B}})-[1-\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{A}) \mathrm{P}(\overline{\mathrm{B}})] \\ &=1-\mathrm{P}(\mathrm{A})+1-\mathrm{P}(\overline{\mathrm{B}})-1+\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})(\overline{\mathrm{B}}) \\ &=1-\mathrm{P}(\mathrm{A}) \mathrm{P}(\overline{\mathrm{B}}) \\ &\Rightarrow \mathrm{P}(\overline{\mathrm{A}} \cup \mathrm{B})=1-\mathrm{P}(\mathrm{A}) \mathrm{P}(\overline{\mathrm{B}})-(1) \end{aligned}$
Also given that $P(\overline{A}\cup B)=1-x-(2)$
Comparing (1) and 2 we get $x=1-P(A)P(\overline{B})$

Probability Exercise Fill in the Blanks Question, Question 2.

Answer: $\frac{5}{12},\frac{1}{3}$
Given:
If A and B are independent events such-that
PA=p,
P(B)=2pand
P( Exactly one of $A_1B=\frac{5}{9}$ ) then
p=?
Hint:
$P(A \cap A B)=P(A) P(B)$ if A and B are independent.
Explanation:
As A and B ane independent events.
So $P(A \cap A B)=P(A) P(B)$ - (1)
Now
P( Exadly one of A and B occurs )= $\frac{5}{9}$
⇒P( only A)+P( onlyB)= $\frac{5}{9}$
$\begin{aligned} &\Rightarrow[\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})]+[\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})]=\frac{5}{9} \\ &\Rightarrow[\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})]+[\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}))]=\frac{5}{9} \\ &\Rightarrow\left[\mathrm{P}(\mathrm{A}) \times(1-\mathrm{P}(\mathrm{B}))+\mathrm{P}(\mathrm{B}) \times[1-\mathrm{P}(\mathrm{A})]=\frac{5}{9}\right. \\ &\Rightarrow \mathrm{p}(1-2 \mathrm{p})+2 \mathrm{p}(1-\mathrm{p})=\frac{5}{9} \\ &\Rightarrow \mathrm{p}-2 \mathrm{p}^{2}+2 \mathrm{p}-2 \mathrm{p}^{2}=\frac{5}{9} \\ &\Rightarrow 3 \mathrm{p}-4 \mathrm{p}^{2}=\frac{5}{9} \\ &\Rightarrow 27 \mathrm{p}-36 \mathrm{p}^{2}=5 \\ &\Rightarrow 36 \mathrm{p}^{2}-27 \mathrm{p}+5=0 \end{aligned}$
So using quadratic formula, $\mu$ .
$\begin{aligned} &p=\frac{-(-27) \pm \sqrt{(-27)^{2}-4 \times 36 \times 5}}{2 \times 36} \\ &=\frac{27 \pm \sqrt{729-720}}{72} \\ &=\frac{27 \pm \sqrt{9}}{72} \\ &=\frac{27 \pm 3}{72} \end{aligned}$
$\begin{aligned} p &=\frac{27+3}{72}, \frac{27-3}{72} \\ &=\frac{30}{72}, \frac{24}{72} \\ &=\frac{5}{12}, \text { or } \frac{1}{3} \end{aligned}$

Probability Exercise Fill in the Blanks Question, Question 3.

Answer: $\frac{10}{9}$
Given:
If A and B are two events such that $P(\overline{A}\cup \overline{B})=\frac{2}{3}$ and
$P(A\cup B)=\frac{5}{9}$
Then $P(\overline{A})+P(\overline{B})= ....$
Hint:
Using $P(\overline{A})=1-P(A)$
Explanation:
Let A and B be the two events.
$\begin{gathered} P(\bar{A} \cup \bar{B})=\frac{2}{3} \\ P(\overline{A \cap B})=\frac{2}{3} \\ \Rightarrow 1-P(A \cap B)=\frac{2}{3} \\ \Rightarrow P(A \cap B)=1-\frac{2}{3} \\ \Rightarrow P(A \cap B)=\frac{1}{3} \\ P(A \cup B)=\frac{5}{9} \text { (given) } \end{gathered}$
Also ,
$\begin{aligned} &\begin{aligned} \mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}}) &=(1-\mathrm{P}(\mathrm{A}))+(1-\mathrm{P}(\mathrm{B})) \\ &=1-\mathrm{P}(\mathrm{A})+1-\mathrm{P}(\mathrm{B}) \end{aligned} \\ &=2-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})] \\ &\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=2-(\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})) \end{aligned}$
Now we know,
$\begin{aligned} &\frac{5}{9}=2-(\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}}))-\frac{1}{3} \\ &\Rightarrow \mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})=2-\frac{5}{9}-\frac{1}{3} \\ &\Rightarrow \mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})=\frac{18-5-3}{9}=\frac{10}{9} \\ &\Rightarrow \mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})=\frac{10}{9} \end{aligned}$

Probability Exercise Fill in the Blanks Question, Question 4.

Answer: $\frac{1}{3}$
Given:
If A and B are thoo events such that
$P(A)=P,$
$P(B)=\frac{1}{3}$ and
$P(A\cup B)=\frac{5}{9}$
Then p=………..
Hint:
Use $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
Explanation:
$\begin{aligned} &P(A)=p \\ &P(B)=\frac{1}{3} \\ &P\left(\frac{A}{B}\right)=P \text { and } \\ &P(A \cup B)=\frac{5}{9} \\ &\therefore P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=p \end{aligned}$
$\begin{aligned} &\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{\mathrm{p}}{3} \\ &\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ &\frac{5}{9}=\mathrm{p}+\frac{1}{3}-\frac{\mathrm{P}}{3} \end{aligned}$
And
$\begin{aligned} &\frac{5}{9}-\frac{1}{3}=\frac{2 p}{3} \\ &p=\frac{2}{9} \times \frac{3}{2}=\frac{1}{3} \end{aligned}$
$p =\frac{1}{3}$

Probability Exercise Fill in the Blanks Question, Question 5.

Answer: Independent
Given:
Let A and B lee ther everts.
If $P\left (\frac{A}{B} \right )=P(A)$ ,
Then A is … of B
Hint:
Using $P\left(\frac{A}{B}\right)=P(A)$
Explanation:
Let A and B le two events.
$\begin{aligned} &\therefore \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} \\ &\Rightarrow \mathrm{P}(\mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} \\ &\Rightarrow \mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \end{aligned}$
So, A is independent of B.

Probability Exercise Fill in the Blanks Question, Question 6.

Answer: $P(A\cup B)$
Given:
Let A and B lu two events. Fof such that
$PA\neq 0,$
$P(B)\neq 1$ and
$\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{1-\mathrm{K}}{\mathrm{P}(\overline{\mathrm{B}})^{\prime}}$
Then k=….
Hint:
$P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
Explanation:
Let A and B are two events
$P\left(\frac{\bar{A}}{\bar{B}}\right)=\frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})}$
$=\frac{\mathrm{P}(\mathrm{AUB})}{\mathrm{P}(\mathrm{B})}=\frac{1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})}{\mathrm{P}(\mathrm{B})}-$
But $P\left(\frac{\overline{\mathrm{A}}}{\overline{\mathrm{B}}}\right)=\frac{1-\mathrm{K}}{\mathrm{P}(\overline{\mathrm{B}})}-(2)$
Comparing (1) and (2)
$K=P(A\cup B)$

Probability Exercise Fill in the Blanks Question, Question 7.

Answer: 0
Given:
If two events A and B are mutually exclusive, then
PAB=………… -
Hint:
You must know about mutually exclusive events
Explanation:
We know
$P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}-(1)$
For mutually exclusive events,
$P(A\cap B)=0$
Put in $(1) P\left (\frac{A}{B} \right )=\frac{O}{P(B)}=0$
$\Rightarrow P\left (\frac{A}{B} \right )=0$

Probability Exercise Fill in the Blanks Question, Question 8.

Answer: 1
Given:
If A and B are tho enends such that $A\subseteqq B$ , then
$P\left ( \frac{A}{B} \right )=....$
Hint:
Using $P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}$
Explanation:
Wit A and B ane two evends.
We are given $A\subseteqq B$
$\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A})$
We know ,
$\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}$
But ,
$\begin{aligned} &\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \\ &\therefore \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A})}=1 \\ &\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=1 \end{aligned}$

Probability Exercise Fill in the Blanks Question, Question 9.

Answer: $\frac{2}{5}$
Given:
If
$\begin{aligned} 4 \mathrm{P}(\mathrm{A}) &=6 \mathrm{P}(\mathrm{B}) \\ &=10 \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ &=1 \end{aligned}$
Then $P\left (\frac{A}{B} \right )=...$
Hint:
$P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
Explanation:
$\begin{aligned} &4 \mathrm{P}(\mathrm{A})=10 \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ &\Rightarrow \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{4}{10} \\ &\Rightarrow \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{4}{10} \\ &=\frac{2}{5} \\ &\Rightarrow \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{2}{5} \end{aligned}$

Probability Exercise Fill in the Blanks Question, Question 10.

Answer: $P(B)-P(A\cap B)$
Given:
If A and B ave two evends then P(A∩B)= $P(A\cap B)=.......$
Hint:
Use formula .
Explanation:
Let A and B ower two events
Then,
$P(\bar{A} \cap B)=P(B)-P(A \cap B)$

Probability Exercise Fill in the Blanks Question, Question 11.

Answer: $P(A\cap B)=P(A)-P(A\cap B)$
Given:
If A and B are turo events, then the probability of occurrence of A only is equal to ... clip_image008

Hint:
Using $P(A\cap B)=P(A)-P(A\cap B)$
Explanation:
Let A and B are two events.
Then P (Only A occurs) $=P(A\cap B)$
$=P(A)-P(A\cap B)$

Probability Exercise Fill in the Blanks Question, Question 12.

Answer: $P(A)+P(B)-2P(A\cap B)$
Given:
If A and B are two everts, then the provability of occurrence of exactly one of A and B is equal to ...
Hint:
$P(A\cap B)=P(A)-P(A\cap B)$
Explanation:
Exactly one of the events of ε;A and B is represented by $A\cap B+A\cap B$
Hence ,
$\begin{aligned} \mathrm{P}(A \cap \bar{B})+P(\bar{A} \cap B) &=[P(A)-P(A \cap B)]+[P(B)-P(A \cap B)] \\ &=P(A)+P(B)-2 P(A \cap B) \end{aligned}$

Probability Exercise Fill in the Blanks Question, Question 13.

Answer: Independent
Given:
For two event A and B, If
$P\left (A \right )=P\left (\frac{A}{B} \right ) =\frac{1}{4}$
and
$P\left (\frac{B}{A} \right )=\frac{1}{2}$
then A and B are …... events.
Hint:
$P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
Explanation:
Let A and B are two wents.
Also $\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=(*)$
We know. $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
Put in (*)
$\begin{aligned} &P(A)=\frac{P(A \cap B)}{P(B)} \\ &\Rightarrow P(A \cap B)=P(A) P(B) \end{aligned}$
$\therefore$ A and B, are independent events.

Probability Exercise Fill in the Blanks Question, Question 14.

Answer: b - c
Given:
Let A and B be the two events for which
$P\left (A \right )=a,$
$P(B)=b,$
$P(A\cap B)=C$ , Then
$P(A\cap B)=......$
Hint:
Using $P(\bar{A} \cap B)=P(B)-P(A \cap B)$
Explanation:
Here
$P\left (A \right )=a,$
$P(B)=b,$
$P(A\cap B)=C$ ,
We know,
$\begin{aligned} \mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B}) &=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ &=\mathrm{b}-\mathrm{C} \end{aligned}$

Probability Exercise Fill in the Blanks Question, Question 15.

Answer: $P(A)+P(B)$
Given:
Ret A₁B,C be pair wise independent events with P(C)>0
and $P(A\cap B\cap C)=0$ . If $P\left(\frac{A \cap B}{C}\right)=1-x$ , then x=…
Hint:
Using $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
Explanation:
Let A,B,C be pair wise independent events.
$\begin{aligned} &\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{C}) \\ &\mathrm{P}(\mathrm{B} \cap \mathrm{C})=\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C})-(*) \end{aligned}$
Given
So $P(A\cap B\cap C)=0$
$\begin{aligned} &\mathrm{P}(\bar{A} \cap \overline{\mathrm{B}} \cap \mathrm{C})=0 \\ &P\left(\frac{\overline{\mathrm{A}} \overline{\mathrm{B}}}{\mathrm{C}}\right)=\frac{\mathrm{P}[(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) \cap \mathrm{C}]}{\mathrm{P}(\mathrm{C})} \\ &=\frac{\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{C})-\mathrm{P}(\mathrm{B} \cap \mathrm{C})+\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})}{\mathrm{P}(\mathrm{C})} \\ &=\frac{\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{B}) \mathrm{P}(\mathrm{C})+0}{\mathrm{P}(\mathrm{C})} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \text { Of }(*)] \end{aligned}$
$\begin{aligned} &=\frac{P(C) \times[1-P(A)-P(B)]}{P(C)} \\ &=1-P(A)-P(B) \\ &=1-[P(A)+P(B)]-(1) \end{aligned}$
Also,
$\mathrm{P}\left(\frac{\mathrm{A} \cap \mathrm{B}}{\mathrm{c}}\right)=1-\mathrm{x}-(2)$
From (1) and (2),
We get $x=P(A)+P(B)$

Probability Exercise Fill in the Blanks Question, Question 16.

Answer : $\frac{ x+y+z}{2}$
Given:
Let A,B,C bo there events such that $P(A\cap B\cap C)=0$
P(Exactly one of A and B occurs) =x,
P (Exactly one of B and C occurs) =y
P (Exactly one of A and C occurs )=z.
Then $P(A\cup B\cup C)= ................$
Hint:
Using the formula of $P(A\cup B\cup C)$
Explanation:
$\begin{aligned} &\mathrm{P}(\mathrm{A} \text { or } \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{x} \\ &\mathrm{P}(\mathrm{B} \text { or } \mathrm{C})=\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{y} \\ &\mathrm{P}(\mathrm{A} \text { or } \mathrm{C})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{C})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{C})=2 \end{aligned}$
Adding these three we get,
$\begin{aligned} &2 \mathrm{P}(\mathrm{A})+2 \mathrm{P}(\mathrm{B})+2 \mathrm{P}(\mathrm{C})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})-2 \mathrm{P}(\mathrm{B} \cap \mathrm{C})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\mathrm{x}+\mathrm{y}+\mathrm{z} \\ &\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})-\mathrm{P}(\mathrm{B} \cap \mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{2}-(1) \end{aligned}$
We know,
$\begin{aligned} &P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(C \cap A) \\ &+P(A \cap B \cap C) \\ &\Rightarrow P(A \cup B \cup C)=\frac{x+y+2}{2}+P(A \cap B \cap C) \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \text { of }(1)] \end{aligned}$
But $P(A\cap B\cap C)=0$
$\therefore P(A \cup B \cup C)=\frac{x+y+2}{2}+0=\frac{x+y+z}{2}$

Probability Exercise Fill in the Blanks Question, Question 17

Answer: 0
Given:
If A and B are two events
then, P (exactly one of A and B occurs )=…
Hint:
$P(A or B)=P(A)+P(B)-2P(A\cap B)$
Explanation:
$\begin{aligned} &P(A \cup B)=P(A \cap B) \\ &\Rightarrow P(A)+P(B)-P(A \cap B)=P(A \cap B) \\ &\Rightarrow P(A)+P(B)-2 P(A \cap B)=0 \end{aligned}$
Also P( exactly one of A and B occurs) = P( A or B)
$\begin{aligned} \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) &=2(\mathrm{P}(\mathrm{A} \cap \mathrm{B})) \\ &=0 \end{aligned}$ $[\because of (1)]$

Students can view the Rd Sharma Class 12th Exercise FBQ online for free by visiting the website. This particular exercise has 17 questions. The concept discussed here is Independent Events, Mutually exclusive, conditional probability, Union and intersection, Bayes theorem, Random variable, Bernoulli Trials, Binomial Distribution, equivalence between the axiomatic theory and the classical theory of Probability in the case of common likely results. Based on this relationship, we get chances of events associated with discrete sample spaces.

The solved examples and exercise problems in Rd Sharma Class 12 Chapter 30 Exercise FBQ will aid in our understanding of the total probability theorem and Bayes theorem covered in the final exercises. RD Sharma solution Therefore, a student must use Class 12th RD Sharma Chapter 30 Exercise FBQ Solutions for the chapter "Probability" to practice and understand these concepts in a structured manner.

The benefits of studying from these solutions are:

The solved examples and exercise problems in Rd Sharma Class 12 Chapter 30 Exercise FBQ will aid in our understanding of the total probability theorem and Bayes theorem covered in the final exercises. Therefore, a student must use Class 12th RD Sharma Chapter 30 Exercise FBQ Solutions for the chapter "Probability" to practice and understand these concepts in a structured manner.

The benefits of studying from these solutions are:

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