RD Sharma Solutions Class 12 Mathematics Chapter 30 VSA

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RD Sharma Solutions Class 12 Mathematics Chapter 30 VSA

RD Sharma Solutions Class 12 Mathematics Chapter 30 VSA

Kuldeep MauryaUpdated on 25 Jan 2022, 04:04 PM IST

Every reputed CBSE institution recommends their students to use the RD Sharma Solution books to clarify their doubts at home. However, even though the teachers provide their best in making the students understand the concepts, they only encounter doubts when they learn the concept. And this takes place a lot many times while doing homework in maths. For example, the sums in the chapter Probability make the students perplexed if they are unclear about the concept. And here is where the RD Sharma Class 12th Chapter 30 VSA comes to the rescue.

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

RD Sharma Class 12 Solutions Chapter30 VSA Probability - Other Exercise

Probability Excercise:VSA

Probablity Exercise very short answer question, question 3

Answer:

The probability that the number has same digit is $\frac{1}{25}$
Hint:
Three digit number cannot start with 0 , and hundredth place can have any of the 4 digits.
Given:
Given digits are 0,2,4,6 and 8 . We have to form a number using digits 0,2,4, 6 and 8 .
Explanation:
Since a three digit number cannot begin with the digit 0. The hundredth place can have any of the 4 digits. Also tens & units place can have all the 5 digits.
So S = total possibilities for 3 -digit numbers
$n(S) = 4 \times 5 \times 5=100$
Let A be the event of three digit numbers having all digits same =4
nA=4
Thus,
$PA=\frac{n(A)}{n(S)}\\ =\frac{4}{100}\\ =\frac{1}{25}$

Probablity Exercise very short answer question, question 4

Answer:

The required probability is $\frac{5}{6^{4}}$
Hint:
$P(Getting\; three\; number\; that\; are\; consecutive) = \frac{Total\; number \;of \;favorable \;outcome}{Total \;number\; of \;sample \;space}$
Given:
Three numbers are chosen at random from 1 to 20.
Explanation:
Let S be a sample space. A cube has 6 faces
Total Possible Outcomes in 5 throws = $6^{5}$
So $n(S)=6^{5}$
Let A be any event of getting 7 is by getting two 2’s and one 3’s
$n(A)=\frac{5_{P_{3}}}{2!}\\ =\frac{5 \times 4\times3\times2\times1}{3\times2\times1}\times\frac{1}{2\times1}\\ =30$
$P(A)=\frac{n(A)}{n(S)}\\ =\frac{30}{6_{5}}\\ =\frac{5}{6^{4}}$
Thus , The required probability is $\frac{5}{6^{4}}$

lease solve RD sharma class 12 chapter 30 Probablity Exercise very short answer question, question 4 maths textbook solutio
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Probablity Exercise very short answer question, question 1

Answer:

The probability that the number is divisible by 5 is $\frac{1}{4}$.
Hint:
Let 'S be set of all number formed by the digits 1,2,3,5 & 'A be the event of getting a number divisible by 5 .
So $P(A)=\frac{n(A)}{n(S)}$
Given:
Given digits a re 1,2,3 & 5
Explanation:
Assume S be set of all number formed by 1,2,3 and 5
A be the event of of getting a number divisible by 5.
n(s)= Number of numbers formed by $1,2,3 \& 5 =4!$
$n(A)=3!$
So $P(A)=\frac{n(A)}{n(S)}=\frac{3!}{4!}=\frac{1}{4}$
This, the probability that the number is divisible by 5 is $\frac{1}{4}$.

Probablity Exercise very short answer question, question 2

Answer:

The probability of getting 4 or 5 on each of the dice simultaneously is $\frac{1}{27}$
Hint:
$P(Getting \;4 \;or\; 5\; on\; each\; of\; the\; dice\; simultaneously ) =\frac{ Total \;number \;of \;favorable\; outcome}{Total\; number\; of \;sample\; space}$
Given:
Three dice are thrown
Explanation:
It is given that ,
S = Three dice are thrown.
n(S)= total number of sample space
= 216
A be an event of getting 4 or 5 on each of the dice simultaneously
$A={445,454,544,455,545,554,555,444]$
So, n(A)=8
At $P(A)=\frac{n(A)}{n(S)}=\frac{8}{216}=\frac{1}{27}$
Thus, required probability is $\frac{1}{27}$
The probability of getting 4 or 5 on each of the dice simultaneously is $\frac{1}{27}$

Probablity Exercise very short answer question, question 5

Answer:

Answer : The required probability is $\frac{18}{20c_{3}}$
Hint:
The possible way of getting 7w by getting two 2’s and one 3’s
Given:
An ordinary cube has 4 plane faces, one face marked 2 and another face marked 3
Explanation:
Probability is given that,
Three numbers are chosen at random from 1 to 20
Total Number of cases = 18
So, the required probability =$\frac{18}{20c_{3}}$
Hence, the probability that they are consecutive is $\frac{18}{20c_{3}}$

Probablity Exercise very short answer question, question 6

Answer:

The required probability is $\frac{1}{132}$
Hint:
Assume six girls as one person
Given:
Six boys and 6 girls sit in a row at random.
Explanation:
Let S be the sample space of arrangement of six boys and six girls.
$n(S) = 12!$
Let A be an event that all girl sit together.

Hence, the required probability is $\frac{1}{132}$

Probablity Exercise very short answer question, question 7

Answer:

The value of P(B) is $\frac{2}{7}$
Hint:
$P(A\cup B')=P(A)+P(B')-P(A\cap B')$ and
If A & B are independent then $P(A\cap B)=P(A).P(B)$
Given
$P(A)=0.3\; \& \;P(A \cup B')=0.8$
Explanation:
It is given that,
$P(A)=0.3 \& \\ P(A\cup B')=0.8$
We know,
$P(A\cup B')= P(A)+ P(B') -P(A\cap B)' \\ \Rightarrow P(A\cup B') =P(A)+P(B')-P(A).P(B')\\ \Rightarrow 0.8 =0.3+P(B') - 0.3 . P(B')\\ \Rightarrow 0.5 =P(B')-0.3 P(B' )\Rightarrow P(B') =\frac{0.5}{0.7}=\frac{5}{7}$ (since, A and B are Independent)
Also, we know,
$P(B)=1-P(B')=1-\frac{5}{7}=\frac{2}{7}$
Thus, P(B)=$\frac{2}{7}$

Probablity Exercise very short answer question, question 8

Answer:

The required probability is $\frac{16}{81}$
Hint:
$Probability \; of \;an \;event = \frac{Total\; number\; of \; favorable\; outcome}{Total\; number\; of\; sample\; space}$
Given:
A die is rolled four times
Explanation:
When a die is rolled ,
S = Total No. of outcomes
= {1,2,3,4,5,6}
n(S) = 6
Let A be an event of getting not less than 2 & not greater than 5.
⇒ A={2,3,4,5}
n(A)=4
$\Rightarrow P(A)=\frac{n(A)}{n(S)}=\frac{4}{6}$
Since a die is rolled four times so, the probability in getting four throws
$=\left (\frac{4}{6} \right ) \times \left (\frac{4}{6} \right ) \times \left (\frac{4}{6} \right ) \times \left (\frac{4}{6} \right )\\ =\frac{16}{81}$
Thus, the required probability is = $\frac{16}{81}$

Probablity Exercise very short answer question, question 9

Answer:

The required expression is $P(A)+ P(B) -2P(A\cap B)$
Hint:
P(exactly one of 2 events occurs) =$P(A\cup B)-P(A\cap B)$
Given:
A and B are two events
Explanation:
It is given that ,
A & B are two events
We need to find expression such that exactly one of two events occurs
$\Rightarrow$ P( Exactly one of 2 everts occurs )
$=P(A\cup B)-P(A\cap B) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ [\because P(A\cup B)=P(A)+P(B)-P(A\cap B)]\\ =P(A)+P(B)-P(A\cap B)-P(A\cap B)\\ =P(A)+P(B)-2P(A\cap B)$
Hence , The required expression is $P(A)+P(B)-2P(A\cap B)$

Probablity Exercise very short answer question, question 10

Answer:

The probability that a number selected at random from the set of members {1,2,3….} is $\frac{1}{25}$
Hint:
Identify number from the set of {1,2,………..100] which are perfect cube & hence find probability.
Given:
Given set of number is {1,2, 3,……….100}
Explanation:
S = {1,2,3………………….100}
n(S) = 100
Let A be the event that the number is cube.
A = {1,8,27,64}
n(A)=4
$P(A) = \frac{n(A)}{n(S)}\\ =\frac{4}{100}\\ =\frac{1}{25}$
Hence the required probability is $\frac{1}{25}$

probablity exercise Very short answer question, question 11

Answer:

Probability that A loses is $\frac{3}{7}$
Hint:
Use the formula,
$P(A) = P(A) + P(B) + P(C)\\ = P(S)\\ = 1$
Given:
$P(A) = 2P(B) \\ P(B) = 2P(C)$
Explanation:
$P(A) = 2P(B)\;\; \& \\ P(B) = 2P(C)$
It is given that
Let A denoted the probability that A wins.
We know,
$P(A)+P(B)+P(C)=1 \\ \Rightarrow P(A)+\frac{P(A)}{2}+\frac{P(A)}{4}=1\\ \Rightarrow7P(A) =4\\ \Rightarrow P(A) =\frac{4}{7 }$
Now, the probability that A losses us $\Rightarrow P(A') =1-P(A)=1-\frac{4}{7 }=\frac{3}{7 }$
Thus, the required probability is $\frac{3}{7 }$

Probablity exercise Very short answer question, question 13

Answer:

Answer : $\frac{1}{4}$
Hint:
$\begin{aligned} &P\left(\frac{B}{A \cup B^{\prime}}\right)=\frac{P\left(B \cap\left(A \cup B^{\prime}\right)\right)}{P\left(A \cup B^{\prime}\right)} \\ \end{aligned}$
Given:
$\begin{aligned} &P\left(A^{\prime}\right)=0.3 \\ &P(B)=0.4 \end{aligned}$ and
$\begin{aligned} P\left(A \cap B^{\prime}\right)=0.5 \end{aligned}$ Then
Explanation:
We have
$\begin{aligned} &P(\bar{A})=0.3, P(B)=0.4\end{aligned}$ and $\begin{aligned}& P(A \cap \bar{B})=0.5 \end{aligned}$
According to bay’s theorem
$\begin{aligned} P\left(\frac{B}{A \cap \bar{B}}\right)=\frac{P(B \cap(\bar{A} \cap \bar{B}))}{P(\bar{A} \cap \bar{B})}\\ =\frac{P(B \cap(\overline{A \cup B}))}{P(\bar{A} \cap \bar{B})} \end{aligned}$
$\begin{aligned} =\frac{P(\bar{B} \cap(\overline{A \cup B}))}{P(\bar{A} \cap \bar{B})}\\ =\frac{P(\bar{B} \cup(A \cup B))}{P(\bar{A} \cap \bar{B})} \end{aligned}$
Now
$\begin{aligned} \bar{B} \cup B=U=\phi \end{aligned}$ Where U=universal set
So
$\begin{aligned} P(\bar{B} \cup(A \cup B))=\phi \end{aligned}$
Therefore $\begin{aligned} P\left(\frac{B}{\bar {A} \cap \bar{B}}\right)=0 \end{aligned}$

Probablity exercise Very short answer question, question 14

Answer:

We can write, $P(A \cap B')=P(A).(1-P(B))$
Hint:
If A & B are independent then $P(A \cap B)=P(A).P(B)$
Given:
A & B are two independent events
Explanation:
It is given that , A & B are independent
We know,
$P(A\cap B')=P(A)-P(A\cap B)$
$=P(A)-P(A).P(B)$ { A & B are independent $P(A \cap B)=P(A).P(B)$ }
$\Rightarrow P(A\cap B^{-})=P(A)(1-P(B))$
Hence,
$P(A\cap B')=P(A).(1-P(B))$

Probablity exercise Very short answer question, question 15

Answer:

The value of $P(A\cup B)=0.75$
Hint:
Using the formula,
$P(B | A) = \frac{P(A\cap B)}{P(A)}$ first find $P(A\cap B)$
Then using the relation, $P(A \cup B)=P(A)+P(B)- P(A\cap B)$,
Find the required value
Given:
P(A) = 0.3
P(B) = 0.6
P(B|A) = 0.5
Explanation:
We know ,
$\begin{aligned} &P(B \mid A)=P(A \cap B) \\ &P(B / A)=\frac{P(A \cap B)}{P(A)} \\ \end{aligned}$
$\begin{aligned} &\Rightarrow 0.5=\frac{P(A \cap B)}{0.3} \\ &\Rightarrow P(A \cap B)=0.5 \times 0.3 \\ &\Rightarrow P(A \cap B)=0.15 \end{aligned}$
We know ,
$\begin{aligned} P(A \cup B) &=P(A)+P(B)-P(A \cap B) \\ &=0.3+0.6-0.15 \\ &=0.75 \end{aligned}$
Hence , $P(A\cup B)=0.75$

Probablity exercise Very short answer question, question 16

Answer:

The required probability is $3p^{2}-3p^{3}$
Hint:
Use the relation ,
P(atleast two of A,B,C occurs) = $P(A\cap B\cap C')+P(A\cap B' \cap C)+P(A'\cap B \cap C)$
Given:
A, B & C are three independent events such that $P(A)=P(B)=P(C)=p$
Explanation:
We have, $P(A)=P(B)=P(C)=p$
$\Rightarrow P(A')=P(B')=P(C')=1-p$
Now,
P(atleast two of A,B,C occurs) = $P(A\cap B\cap C')+P(A\cap B' \cap C)+P(A'\cap B \cap C)$
$= P(A).P(B).P(C')+P(A).P(B').P(C) +P(A').P(B).P(C)$
( $\therefore$A , B & C are independent )
$=p.p(1-p)+p(1-p).p+(p-1)p.p\\ =p^{2}(1-p)+p^{2}(1-p)+p^{2}(1-p)\\ =(1-p) 3p^{2}\\ = 3p^{2}- 3p^{3}$

Probablity exercise Very short answer question, question 17

Answer:

The required expression is $P(A)+P(B)-2P(A).P(B)$
Hint:
If A and B are independent then $P(A\cap B)=P(A).P(B)$ and
P(exactly one of the two events occurs) = $P(A \cup B)-P(A \cap B)$
Given:
A and B are independent events
Explanation:
It is given that ,
A and B are independent events
Now ,
P(exactly one of the two events occurs) = $P(A \cup B)-P(A \cap B)$
$=P(A)+P(B)-P(A\cap B)-P(A\cap B)$
$=P(A)+P(B)-P(A).P(B)-P(A).P(B)$ ($\therefore$ A , B are independent)
$=P(A)+P(B)-2P(A).P(B)$
Hence, the required expression is $P(A)+P(B)-2P(A).P(B)$

Probablity exercise Very short answer question, question 18

Answer:

The value of p is $\frac{5}{12}$ or $\frac{1}{3}$
Hint:
If A and B are independent then,
$P(A\cap B)=P(A).P(B)$ and
P(exactly one of the two events occurs) = $P(A)+P(B) - 2P(A \cap B)$
Given:
A and B are independent events such that
P(A) = p
P(B) = 2p
P(exactly one of A and B occurs) = $\frac{5}{9}$
Explanation:
It is given that
P(exactly one of A and B occurs) = $\frac{5}{9}$
$\begin{aligned} &\Rightarrow \frac{5}{9}=P(A)+P(B)-2 P(A \cap B) \\ &=p+2 p-2 \cdot P(A) \cdot P(B) \\ \end{aligned}$ [ since,PA ∩B=PAPB]

⇒$\begin{aligned} &\Rightarrow \frac{5}{9}=p+2 p-2 \cdot p \cdot 2 p \\ &\Rightarrow \frac{5}{9}=3 p-4 p^{2} \\ &\Rightarrow 27 p-36 p^{2}=5 \\ &\Rightarrow 36 p^{2}-27 p+5=0 \\ \end{aligned}$
Using quadratic formula, we have
$\begin{aligned} p &=-\frac{(-27) \pm \sqrt{(-27)^{2}-4(36) 5}}{2 \times 36} \\ \end{aligned}$
$\begin{aligned} &=\frac{27 \pm \sqrt{729-720}}{72} \\ &=\frac{27 \pm \sqrt{9}}{72} \\ &=\frac{27 \pm 3}{72} \\ \end{aligned}$
$\begin{aligned} &\Rightarrow p=\frac{27+3}{72} \text { or } \quad p=\frac{27-3}{72} \\ &\Rightarrow p=\frac{30}{72} \quad \text { or } p=\frac{24}{72}\\ &\Rightarrow p=\frac{5}{12} \quad \text { or } p=\frac{1}{3} \end{aligned}$
Hence, the value of p is $\frac{5}{12}$ or $\frac{1}{3}$

Chapter 30, Probability, has seven exercises in total, ex 30.1 to ex 30.7. However, when it comes to the Very Short Answer (VSA) part, most students get confused on how to solve quickly. There are eighteen questions in the VSA section. The concepts included here are the probability of dice, divisibility, coins, cards, the occurrence of two events based on Union and intersection, with replacement and without replacement, binomial, random variable and mutually exclusive event. If the students struggle to guess the exact formula used in these concepts, they can refer to the RD Sharma Class 12 Solutions Chapter 30 VSA book.

One challenging task in solving the VSA part is that the solutions must be short and direct. Therefore, the usage of tricks and shortcuts is highly encouraged. RD Sharma Solutions If the students are unaware of it, they spend much time in this portion during exams. The RD Sharma Class 12th Chapter 30 VSA solution material provides the students with various shortcuts to solve the probability sums. As it follows the NCERT syllabus, the CBSE students feel it is easy to cross their benchmark scores.

The Class 12 RD Sharma Chapter 30 VSA Solution material solutions are provided and verified by experts in the respective domain. Various practice questions make the students used to the VSA type questions. This eventually makes them save time during the examination, which they can spend on rechecking the answers and correcting their mistakes.

The students can use the RD Sharma Class 12 Solutions Probability Chapter 30 VSA book while they encounter doubts in their homework assignments. The Career360 website gives access for everyone to view and download the RD Sharma Class 12th Chapter 30 VSA reference material for free of cost.

The questions for the public exams are also taken from the RD Sharma books. Therefore, the RD Sharma Class 12 Solutions Chapter 30 FBQ helps the students prepare for their public exams effortlessly. Various students have benefitted by using these books for their public exams. So, download your copy of RD Sharma books and commence the preparation for the public exams.