RD Sharma Solutions Class 12 Mathematics Chapter 3 CSBQ

RD Sharma Solutions Class 12 Mathematics Chapter 30 CSBQ

Edited By Kuldeep Maurya | Updated on Jan 25, 2022 04:04 PM IST

The Majority of CBSE schools prescribe the RD Sharma solution books to their students. This is because the teacher or a tutor cannot be with the students 24 x 7. Whenever the students study or do their homework by themselves, they encounter a lot of doubts. Primarily when they work out maths homework on the Case-Based Questions (CSBQ) in the Probability chapter, the complexities know no limits. Here is where the RD Sharma Class 12th Chapter 30 CSBQ solution book lends a helping hand.

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RD Sharma Class 12 Solutions Chapter30 CSBQ Probability - Other Exercise

Probability Excercise:CSBQ

Probability Exercise Case Study Based Question, question 1 (ii)

Answer:
(c)
Hint:
You must know rules of conditional probability
Given:
Vinay process 50% forms, Sonia process 20% and Iqbal process 30%. Vinay has an error of 0.06, Sonia has error rate 0.04 and Iqbal has error rate of 0.03
Solution:
Probability that Sonia processed the form and committed an error
$\begin{aligned} &=20 \% \times 0.04 \\ &=\frac{20}{100} \times 0.04 \\ &=0.008 \end{aligned}$
So, (c) is correct answer

Probability Exercise Case Study Based Question, question 1 (iv)

Answer:
(d)
Hint:
You must know rules of conditional probability
Given:
Vinay process 50% forms, Sonia process 20% and Iqbal process 30%. Vinay has an error of 0.06, Sonia has error rate 0.04 and Iqbal has error rate of 0.03
Solution:
First we find,
The probability that an error is committed in processing given that Vinay processed form
i.e. $P\left (\frac{V}{E} \right )$
Now,
$\begin{aligned} &P\left(\frac{V}{E}\right)=\frac{P(V) \cdot P\left(\frac{E}{V}\right)}{P(V) \cdot P\left(\frac{E}{V}\right)+P(S) \cdot P\left(\frac{E}{S}\right)+P(I) \cdot P\left(\frac{E}{I}\right)} \\ &=\frac{50 \% \times 0.06}{50 \% \times 0.06+20 \% \times 0.04+30 \% \times 0.03} \\ &=\frac{\frac{50}{100} \times 0.06}{\frac{50}{100} \times 0.06+\frac{20}{100} \times 0.04+\frac{30}{100} \times 0.03} \\ &=\frac{0.30}{0.30+0.08+0.09} \\ &=\frac{0.30}{0.47} \\ &=\frac{30}{47} \end{aligned}$
Now,
Probability form has an error but not processed by Vinay
$\begin{aligned} &=1-P\left(\frac{V}{E}\right) \\ &=1-\frac{30}{47} \end{aligned} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[P\left(\frac{V}{E}\right)=\frac{30}{47}\right]$
$\begin{aligned} &=\frac{47-30}{47} \\ &=\frac{17}{47} \end{aligned}$
So, (d) is correct answer

Probability Exercise Case Study Based Question, question 3 (iii).

Answer:
(d)
Hint:
You must know rules of conditional probability
Given:
90% test detect the disease, 10% go undetected, 99% judged HIV –ve and 1% judged
HIV +ve, 0.1% have HIV
Solution:
Let the events be,
E=Person selected has HIV
F=Person selected does not have HIV
G=Test judges HIV +ve
$\begin{aligned} &P(G)=P(E) \cdot P\left(\frac{G}{E}\right)+P(F) \cdot P\left(\frac{G}{F}\right) \\ &=0.001 \times 0.9+0.999 \times 0.01 \\ &=9 \times 10^{-4}+99.9 \times 10^{-4} \\ &=\frac{9+99.9}{10^{4}} \\ &=\frac{108.9}{10^{4}} \\ &=\frac{1089}{10^{5}} \\ &=0.01089 \end{aligned}$
So, (d) is correct answer

Probability Exercise Case Study Based Question, question 4 (v).

Answer:
(d)
Hint:
You must know rules of finding probability
Given:
Aarushi solve the problem correctly $\frac{1}{3}=P(A)$
Avni solve the problem correctly $\frac{2}{7}=P(B)$
Mira solve the problem correctly $\frac{3}{8}=P(C)$
Solution:
Probability that none of them solves the problem correctly
$\begin{aligned} &P(\bar{A}) \cdot P(\bar{B}) \cdot P(\bar{C})=(1-P(A)) \cdot(1-P(B)) \cdot(1-P(C)) \\ &=\left(1-\frac{1}{3}\right) \cdot\left(1-\frac{2}{7}\right) \cdot\left(1-\frac{3}{8}\right) \\ &=\left(\frac{2}{3}\right) \cdot\left(\frac{5}{7}\right) \cdot\left(\frac{5}{8}\right) \\ &=\frac{25}{84} \end{aligned}$
So, (d) is correct answer

Probability Exercise Case Study Based Question, question 1 (i)

Answer:
(b)
Hint:
You must know rules of conditional probability
Given:
Vinay process 50% forms, Sonia process 20% and Iqbal process 30%. Vinay has an error of 0.06, Sonia has error rate 0.04 and Iqbal has error rate of 0.03
Solution:
The conditional probability that an error is committed in processing given that Sonia processed the form
i.e. $P\left (\frac{E}{S} \right )$
Sonia’s error rate = 0.04
$P\left (\frac{E}{S} \right )=0.4$
So, (b) is correct option

Probability Exercise Case Study Based Question, question 1 (i)

Probability Exercise Case Study Based Question, question 1 (iii)

Answer:
(b)
Hint:
You must know rules of solving probability
Given:
Vinay process 50% forms, Sonia process 20% and Iqbal process 30%. Vinay has an error of 0.06, Sonia has error rate 0.04 and Iqbal has error rate of 0.03
Solution:
The total probability of committing an error = Probability Vinay processed form x Vinay’s error rate + probability Sonia processed form

Sonia’s error rate + probability Iqbal processed form x Iqbal error rate
$\begin{aligned} &=50 \% \times 0.06+20 \% \times 0.04+30 \% \times 0.03 \\ &=\frac{50}{100} \times 0.06+\frac{20}{100} \times 0.04+\frac{30}{100} \times 0.03 \\ &=0.030+0.008+0.009 \\ &=0.047 \end{aligned}$
So (b) is correct answer

Probability Exercise Case Study Based Question, question 1 (v)

Answer:
(d)
Hint:
You must know rules of conditional probability
Given:
Vinay process 50% forms, Sonia process 20% and Iqbal process 30%. Vinay has an error of 0.06, Sonia has error rate 0.04 and Iqbal has error rate of 0.03
Solution:
Now
E₁ = Vinay
E₂ = Sonia
E₃ = Iqbal
Now
$\begin{aligned} &\sum_{i=1}^{3} P\left(\frac{E_{i}}{A}\right)=P\left(\frac{E_{1}}{A}\right)+P\left(\frac{E_{2}}{A}\right)+P\left(\frac{E_{3}}{A}\right) \\ &=\frac{P\left(E_{1}\right) \cdot P\left(\frac{A}{E_{1}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{A}{E_{2}}\right)+P\left(E_{3}\right) P\left(\frac{A}{E_{3}}\right)} \\ &+\frac{P\left(E_{2}\right) \cdot P\left(\frac{A}{E_{2}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{A}{E_{2}}\right)+P\left(E_{3}\right) P\left(\frac{A}{E_{3}}\right)} \\ &+\frac{P\left(E_{3}\right) \cdot P\left(\frac{A}{E_{3}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{A}{E_{2}}\right)+P\left(E_{3}\right) P\left(\frac{A}{E_{3}}\right)} \end{aligned}$
$\begin{aligned} &=\frac{\left(P\left(E_{1}\right) \cdot P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{A}{E_{2}}\right)+P\left(E_{3}\right) \cdot P\left(\frac{A}{E_{3}}\right)\right)}{\left(P\left(E_{1}\right) \cdot P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{A}{E_{2}}\right)+P\left(E_{3}\right) \cdot P\left(\frac{A}{E_{3}}\right)\right)} \\ \end{aligned}$
$=1$
So, (d) is correct answer

Probability Exercise Case Study Based Question, question 2 (i)

Answer:
(d)
Hint:
You must know rules of conditional probability
Given:
6% people are left handed with blood group O, and 10% of those with other blood group are left handed 30% of the people have blood group O
Solution:
Probability of selecting a left handed person given that he /she has blood group O
$E_1=$ The person selected is of blood group O
$E_2=$ The person selected is of other than blood group O
$E_3=$ Selected person is left handed
$\begin{aligned} &P\left(E_{1}\right)=0.3 \\ &P\left(E_{2}\right)=0.7 \\ &P\left(\frac{E}{E_{1}}\right)=0.06 \\ &P\left(\frac{E}{E_{2}}\right)=0.1 \end{aligned}$
P (Left handed person having blood group O)
$=P\left(\frac{E}{E_{1}}\right)=\frac{6}{100}=0.06$
So, (d) is correct option

Probability Exercise Case Study Based Question, question 2 (ii)

Answer:
(b)
Hint:
You must know rules of conditional probability
Given:
People blood group O left handed 6% and people blood group O 30% and no blood group O but left handed 10%
Solution:
Probability of choosing left handed person given that he / she does not have blood group O
$P\left(\frac{E}{E_{2}}\right)=\frac{10}{100}=0.1$

Probability Exercise Case Study Based Question, question 2 (iii)

Answer:
(a)
Hint:
You must know rules of conditional probability
Given:
6% people are left handed with blood group O, 10% of those with other blood group are left handed, 30% of people have blood group O
Solution:
$E_1=$ The people selected is of blood group O
$E_2=$ The people selected is of other than blood group O
$E_3=$ Selected person is left handed
$\begin{aligned} &P\left(E_{1}\right)=0.3 \\ &P\left(E_{2}\right)=0.7 \\ &P\left(\frac{E}{E_{1}}\right)=0.06 \\ &P\left(\frac{E}{E_{2}}\right)=0.1 \end{aligned}$
Probability of selecting a left handed person
$\begin{aligned} &P(E)=P\left(\frac{E}{E_{1}}\right) \cdot P\left(E_{1}\right)+P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right) \\ &=0.3 \times 0.06+0.7 \times 0.1 \\ &=0.018+0.07 \\ &=0.088 \end{aligned}$
So, (a) is correct answer

Probability Exercise Case Study Based Question, question 2 (iv)

Answer:
(b)
Hint:
You must know rules of conditional probability
Given:
6% people are left handed with blood group O, 10% of those with other blood group are left handed, 30% of people have blood group O
Solution:
$E_1=$ The people selected is of blood group O
$E_2=$ The people selected is of other than blood group O
$E=$ Selected person is left handed
$\begin{aligned} &P\left(E_{1}\right)=0.3 \\ &P\left(E_{2}\right)=0.7 \\ &P\left(\frac{E}{E_{1}}\right)=0.06 \\ &P\left(\frac{E}{E_{2}}\right)=0.1 \end{aligned}$
If person is selected at random, probability that he/she has blood group O
Using Baye’s theorem,
$P\left(\frac{E_{1}}{E}\right)=\frac{P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}$
$\begin{aligned} &=\frac{0.3 \times 0.06}{0.03 \times 0.06+0.7 \times 0.1} \\ &=\frac{0.018}{0.018 \times 0.07} \\ &=\frac{9}{44} \end{aligned}$
So, (b) is correct answer

Probability Exercise Case Study Based Question, question 2 (v)

Answer:
(c)
Hint:
You must know rules of conditional probability and baye’s theorem,
Given:
6% people are left handed with blood group O, 10% of those with other blood group are left handed, 30% of people have blood group O
Solution:
Probability that a randomly selected person is right handed
That person is not left handed $P(\bar{E})=1-P(E)$
$\begin{aligned} &P(\bar{E})=1-P(E) \\ &=1-\left[P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)\right] \\ &=1-[0.3 \times 0.06+0.7 \times 0.1] \\ &=1-[0.018+0.07] \\ &=1-0.088 \\ &=1-\frac{88}{1000}=\frac{912}{1000}=\frac{114}{125} \end{aligned}$
So, (c) is correct answer

Probability Exercise Case Study Based Question, question 3 (i).

Answer:
(b)
Hint:
You must know rules of finding probability
Given:
90% test detect the disease, 10% go undetected, 99% judged HIV –ve and 1% judged
HIV +ve, 0.1% have HIV
Solution:
Let the events ,
E =Person selected has HIV
F= Person selected does not have HIV
G= Test judges HIV +ve
$\begin{aligned} &P(E)=0.001 \\ &P(F)=0.999 \\ &P\left(\frac{G}{E}\right)=0.9 \\ &P\left(\frac{G}{F}\right)=0.01 \end{aligned}$
Probability the person tested HIV +ve given that he/she actually has HIV is,
$\begin{aligned} &P\left(\frac{G}{E}\right)=90 \% \\ &=\frac{90}{100}=0.9 \end{aligned}$
So, (b) is the correct answer

Probability Exercise Case Study Based Question, question 3 (ii).

Answer:
(c)
Hint:
You must know rules of finding probability
Given:
90% test detect the disease, 10% go undetected, 99% judged HIV –ve and 1% judged
HIV +ve, 0.1% have HIV
Solution:
E= Person selected has HIV
F= Person selected does not have HIV
G= Test judges HIV +ve
Probability that person tested HIV +ve give that he/she actually is not HIV infected,
$\begin{aligned} &P\left(\frac{G}{F}\right)=1 \% \\ &=\frac{1}{100}=0.01 \end{aligned}$

Probability Exercise Case Study Based Question, question 3 (iv).

Answer:
(c)
Hint:
You must know rules of conditionalprobability
Given:
90% test detect the disease, 10% go undetected, 99% judged HIV –ve and 1% judged
HIV +ve, 0.1% have HIV
Solution:
E= Person selected has HIV
F= Person selected does not have HIV
G= Test judges HIV +ve
$\begin{aligned} &P(E)=0.001 \\ &P(F)=0.999 \\ &P\left(\frac{G}{E}\right)=0.9 \\ &P\left(\frac{G}{F}\right)=0.01 \end{aligned}$
We need to find the probability that person selected actually has HIV
Using baye’s theorem,
$\begin{aligned} &P\left(\frac{E}{G}\right)=\frac{P(E) \cdot P\left(\frac{G}{E}\right)}{P(E) \cdot P\left(\frac{G}{E}\right)+P(F) \cdot P\left(\frac{G}{F}\right)} \\ &=\frac{0.001 \times 0.9}{0.001 \times 0.9+0.999 \times 0.01} \\ &=\frac{9 \times 10^{-4}}{9 \times 10^{-4}+99.9 \times 10^{-4}} \\ &=\frac{10^{-4} \times 9}{10^{-4}(9+99.9)} \\ &=\frac{9}{108.9}=\frac{90}{1089} \end{aligned}$
So, (c) is correct answer

Probability Exercise Case Study Based Question, question 3 (v).

Answer:
(c)
Hint:
You must know rules of conditional probability
Given:
90% test detect the disease, 10% go undetected, 99% judged HIV –ve and 1% judged
HIV +ve, 0.1% have HIV
Solution:
The probability that the person selected is not having HIV given that the test report diagnosed is +ve
It means, the person selected is not actually has HIV
$\begin{aligned} &P\left(\frac{\bar{E}}{G}\right)=1-P\left(\frac{E}{G}\right) \\ &=1-\frac{90}{1089} \\ &=\frac{1089-90}{1089}=\frac{999}{1089} \end{aligned}$
So, (c) is correct answer

Probability Exercise Case Study Based Question, question 4 (i).

Answer:
(b)
Hint:
You must know rules of finding probability
Given:
Aarushi solve the problem correctly $\frac{1}{3}=P(A)$
Avni solve the problem correctly $\frac{2}{7}=P(B)$
Mira solve the problem correctly $\frac{3}{8}=P(C)$
Solution:
Probability the all solve the problem correctly
$\begin{aligned} &=P(A) \cdot P(B) \cdot P(C) \\ &=\frac{1}{3} \times \frac{2}{7} \times \frac{3}{8} \\ &=\frac{1}{28} \end{aligned}$
So, (b) is correct answer

Probability Exercise Case Study Based Question, question 4 (ii).

Probability Exercise Case Study Based Question, question 4 (iii).

Answer:
(a)
Hint:
You must know rules of finding probability
Given:

Aarushi solve the problem correctly $\frac{1}{3}=P(A)$
Avni solve the problem correctly $\frac{2}{7}=P(B)$
Mira solve the problem correctly $\frac{3}{8}=P(C)$
Solution:
Probability that only Aarushi and Mira solve the problem correctly ,
$\begin{aligned} &=P(A) \cdot P(\bar{B}) \cdot P(\bar{C})-P(A) \cdot P(C) \cdot P(\bar{B}) \\ &=P(A) \cdot P(\bar{B}) \cdot[P(\bar{C})-P(C)] \\ &=\left(\frac{1}{3}\right) \cdot\left(\frac{5}{7}\right) \cdot\left[\frac{5}{8}-\frac{3}{8}\right] \\ &=\frac{5}{21} \cdot\left(\frac{2}{8}\right)=\frac{5}{21} \cdot\left(\frac{1}{4}\right) \\ &=\frac{5}{84} \end{aligned}$
So, (a) is the correct answer

Probability Exercise Case Study Based Question, question 4 (iv).

Answer:
(c)
Hint:
You must know rules of finding probability
Given:
Aarushi solve the problem correctly $\frac{1}{3}=P(A)$
Avni solve the problem correctly $\frac{2}{7}=P(B)$
Mira solve the problem correctly $\frac{3}{8}=P(C)$
Solution:
Probability that exactly one of them solve the problem correctly
$\begin{aligned} &=P(A) \cdot P(\bar{B}) \cdot P(\bar{C})+P(\bar{A}) \cdot P(B) \cdot P(\bar{C})+P(\bar{A}) \cdot P(\bar{B}) \cdot P(C) \\ &=\left(\frac{1}{3}\right) \cdot\left(\frac{5}{7}\right) \cdot\left(\frac{5}{8}\right)+\left(\frac{2}{3}\right) \cdot\left(\frac{2}{7}\right) \cdot\left(\frac{5}{8}\right)+\left(\frac{2}{3}\right) \cdot\left(\frac{5}{7}\right) \cdot\left(\frac{3}{8}\right) \\ &=\frac{25+20+30}{168}=\frac{75}{168} \\ &=\frac{25}{56} \end{aligned}$
So, (c) is correct answer

Probability Exercise Case Study Based Question, question 5 (i).

Answer:
(b)
Hint:
You must know rules of finding probability
Given:
Machine A produce 1% defective item, B produce 5%, C produce 7%. Operator A on job for 50% of time, B on job for 30% and C on job for 20%
Solution:
A= Event that item produced by operator A
B= Item produced by operator B
C= Item produced by operator C
D= Item produced is defective
An item is chosen from the items produced, the probability that it is defective is
$\begin{aligned} &P(D)=P(A) \cdot P\left(\frac{D}{A}\right)+P(B) \cdot P\left(\frac{D}{B}\right)+P(C) \cdot P\left(\frac{D}{C}\right) \\ &=50 \% \times 1 \%+30 \% \times 5 \%+20 \% \times 7 \% \\ &=0.5 \times 0.01+0.3 \times 0.05+0.2 \times 0.07 \\ &=0.034 \\ &=\frac{34}{1000} \\ &=\frac{17}{500} \end{aligned}$
So, (b) is correct answer

Probability Exercise Case Study Based Question, question 5 (ii).

Answer:
(c)
Hint:
You must know rules of finding probability
Given:
Machine A produce 1% defective item, B produce 5%, C produce 7%. Operator A on job for 50% of time B on job for 30% and C on job for 20%
Solution:
The probability that an item produced is defective given that it is produced by C
$\begin{aligned} &P\left(\frac{D}{C}\right)=7 \% \\ &=\frac{7}{100} \\ &=0.07 \end{aligned}$
So, (c) is correct answer

Probability Exercise Case Study Based Question, question 5 (iii).

Answer:
(a)
Hint:
You must know rules of finding probability
Given:
Machine A produce 1% defective item, B produce 5%, C produce 7%. Operator A on job for 50% of time B on job for 30% and C on job for 20%
Solution:
A= Event that item produced by operator A
B= Item produced by operator B
C= Item produced by operator C
D= Item produced is defective
$\begin{aligned} &P(A)=0.5, P(B)=0.3, P(C)=0.2 \\ &P\left(\frac{D}{A}\right)=0.01, P\left(\frac{D}{B}\right)=0.05, P\left(\frac{D}{C}\right)=0.07 \end{aligned}$
A defective item is produced, probability that it was produced by A,
Using Baye’s theorem
$\begin{aligned} &P\left(\frac{A}{D}\right)=\frac{P(A) \cdot P\left(\frac{D}{A}\right)}{P(A) \cdot P\left(\frac{D}{A}\right)+P(B) \cdot P\left(\frac{D}{B}\right)+P(C) \cdot P\left(\frac{D}{C}\right)} \\ &=\frac{0.5 \times 0.01}{0.5 \times 0.01+0.3 \times 0.05+0.2 \times 0.07} \\ &=\frac{0.005}{0.005+0.015+0.014} \\ &=\frac{0.005}{0.034} \\ &=\frac{5}{34} \end{aligned}$
So, (a) is correct answer

Probability Exercise Case Study Based Question, question 5 (iv).

Answer:
(b)
Hint:
You must know rules of finding probability
Given:
Machine A produce 1% defective item, B produce 5%, C produce 7%. Operator A on job for 50% of time B on job for 30% and C on job for 20%
Solution:
A= Event that item produced by operator A
B= Item produced by operator B
C= Item produced by operator C
D=Item produced is defective
$\begin{aligned} &P(A)=0.5, P(B)=0.3, P(C)=0.2 \\ &P\left(\frac{D}{A}\right)=0.01, P\left(\frac{D}{B}\right)=0.05, P\left(\frac{D}{C}\right)=0.07 \end{aligned}$
Probability of defective item is produced by B
Using Baye’s theorem,
$\begin{aligned} &P\left(\frac{B}{D}\right)=\frac{P(B) \cdot P\left(\frac{D}{B}\right)}{P(A) \cdot P\left(\frac{D}{A}\right)+P(B) \cdot P\left(\frac{D}{B}\right)+P(C) \cdot P\left(\frac{D}{C}\right)} \\ &=\frac{0.3 \times 0.05}{0.5 \times 0.01+0.3 \times 0.05+0.2 \times 0.07} \\ &=\frac{0.015}{0.005+0.015+0.014} \\ &=\frac{0.015}{0.034} \\ &=\frac{15}{34} \end{aligned}$
So, (b) is correct answer

Probability Exercise Case Study Based Question, question 5 (v).

Answer:
(d)
Hint:
You must know rules of finding probability
Given:
Machine A produce 1% defective item, B produce 5%, C produce 7%. Operator A on job for 50% of time B on job for 30% and C on job for 20%
Solution:
$\begin{aligned} &P(A)=0.5, P(B)=0.3, P(C)=0.2 \\ &P\left(\frac{D}{A}\right)=0.01, P\left(\frac{D}{B}\right)=0.05, P\left(\frac{D}{C}\right)=0.07 \end{aligned}$
A= Event that item produced by operator A
B= Item produced by operator B
C= Item produced by operator C
D= Item produced is defective
Defective item is produced, that the probability it was produced by B or C
$P\left(\frac{B}{D}\right)+P\left(\frac{C}{D}\right)$
$=\frac{P(B) \cdot P\left(\frac{D}{B}\right)}{P(A) \cdot P\left(\frac{D}{A}\right)+P(B) \cdot P\left(\frac{D}{B}\right)+P(C) \cdot P\left(\frac{D}{c}\right)}+\frac{P(C) \cdot P\left(\frac{D}{C}\right)}{P(A) \cdot P\left(\frac{D}{A}\right)+P(B) \cdot P\left(\frac{D}{B}\right)+P(C) \cdot P\left(\frac{D}{C}\right)}$
$\begin{aligned} &=\frac{0.3 \times 0.05}{0.5 \times 0.01+0.3 \times 0.05+0.2 \times 0.07}+\frac{0.2\times 0.07}{0.5 \times 0.01+0.3 \times 0.05+0.2 \times 0.07} \\ &=\frac{0.015}{0.034}+\frac{0.014}{0.034} \\ &=\frac{0.029}{0.034} \\ &=\frac{29}{34} \end{aligned}$
So, (d) is correct answer

Probability Exercise Case Study Based Question, question 6 (i).

Answer:
(c)
Hint:
You must know rules of finding probability
Given:
Chance of question solved by Aarushi is 30%, Avni is 25% and Mira is 45%, the probability error done by Aarushi 0.01, Avni 0.012, Mira 0.02
Solution:
The probability that an error is done in solving given that tried by Mira
P (Mira committed an error) = 0.02

Probability Exercise Case Study Based Question, question 6 (ii).

Answer:
(b)
Hint:
You must know rules of conditional probability
Given:
Chance of question solved by Aarushi is 30%, Avni is 25% and Mira is 45%, the probability error done by Aarushi 0.01, Avni 0.012, Mira 0.02
Solution:
The probability that Mira solved the question and committed an error
P (Mira solved and committed error) = $45\%\times 0.02$
$\begin{aligned} &=\frac{45}{100} \times 0.02 \\ &=0.45 \times 0.02 \\ &=0.0090 \end{aligned}$
So, (b) is correct answer

Probability Exercise Case Study Based Question, question 6 (iii).

Answer:
(a)
Hint:
You must know rules of finding probability
Given:
Chance of question solved by Aarushi is 30%, Avni is 25% and Mira is 45%, the probability error done by Aarushi 0.01, Avni 0.012, Mira 0.02
Solution:
The total probability of committing error in solving question,
P (Aarsuhi) + P (Avni) + P (Mira)
$\begin{aligned} &=30 \% \times 0.01+25 \% \times 0.012+45 \% \times 0.02 \\ &=\frac{30}{100} \times 0.01+\frac{25}{100} \times 0.012+\frac{45}{100} \times 0.02 \\ &=0.003+0.003+0.009 \\ &=0.015 \end{aligned}$
So, (a) is correct answer

Probability Exercise Case Study Based Question, question 6 (iv).

Answer:
(b)
Hint:
You must know rules of finding probability
Given:
Chance of question solved by Aarushi is 30%, Avni is 25% and Mira is 45%, the probability error done by Aarushi 0.01, Avni 0.012, Mira 0.02
Solution:
P (Error is not done by Aarushi) = 1 -P (Aarush)
=1 - 0.01
=0.99
So, (b) is correct answer

Probability Exercise Case Study Based Question, question 6 (v).

Answer:
(a)
Hint:
You must know rules of conditional probability
Given:
Chance of question solved by Aarushi is 30%, Avni is 25% and Mira is 45%, the probability error done by Aarushi 0.01, Avni 0.012, Mira 0.02
Solution:
If an error is noticed in the solution, the probability that it was solved by Avni or Mira
By Baye’s theorem,
$P(\text { Avni })+P(\text { Mira })=\frac{P(\text { Avni })+P(\text { Mira })}{P(\text { Aarushi })+P(\text { Avni })+P(\text { Mira })}$
$\begin{aligned} &=\frac{0.25 \times 0.012+0.45 \times 0.02}{0.30 \times 0.01+0.25 \times 0.012+0.45 \times 0.02} \\ &=\frac{0.003+0.009}{0.003+0.003+0.009} \\ &=\frac{0.012}{0.015} \\ &=0.8 \end{aligned}$

So, (a) is correct answer

The grade 12 portions for maths in chapter 30 consists of seven exercises, ex 30.1 to ex 30.7. When it comes to the CSBQ part, there are 30 questions, including the subparts. The concept in these questions is conditional Probability, error rate probability, random selection, union and intersection. The students find it hard to attend the mathematics CSBQ while doing their homework. And when it is asked in the Probability chapter, they get even more confused. The presence of RD Sharma Class 12 Chapter 30 CSBQ reference material is undeniable.

The questions present in this portion are tricky and require good clarity in the concept. RS Sharma solutions Only when a student has a strong foundation in Probability from the previous exercises will they be able to solve this section. Then, they can use the RD Sharma Class 12th Chapter 30 CSBQ solution material to refer to the answers. As this book follows the NCERT pattern, every section related to the CBSE syllabus is given.

The Case-Based questions develop a deeper understanding of the concept as it is based on real-life scenarios. The solutions for these questions are provided by the staff members who have expertise in Probability. In addition, there are many practice questions in the Class 12 RD Sharma Chapter 30 CSBQ Solution that help them learn Probability in-depth.

The students are asked to do this part as their assignments, for which they can refer to the RD Sharma Class 12 Solutions Probability Chapter 30 CSBQ. This reference material is freely available on the Career360 website. No one is required to pay money or any other monetary charge to view or download the RD Sharma Class 12th Chapter 30 CSBQ book.

Most of the previous batch students have benefitted by using these books for their exam preparation. You can also expect questions from the RD Sharma Class 12 Solutions Chapter 30 CSBQ in your public exams. Therefore, it is a wise choice to use this material from day one of preparation. Visit the Career 360 website to download your set of solution materials.

RD Sharma Chapter wise Solutions

Frequently Asked Question (FAQs)

1. Where can the class 12 students refer to clarify their doubts in mathematics Case-Based Questions?

The students can refer to the RD Sharma Class 12th Chapter 30 CSBQ reference material to work out this portion effortlessly.

2. How many CBSQ questions are given in the Probability chapter?

There are six Cases in which each case has five questions. Therefore, 30 questions are asked in the Probability chapter, and the students can use the RD Sharma Class 12th Chapter 30 CSBQ book to solve these sums.

3. Are the solutions given in the RD Sharma books verified?

All the answers provided in the RD Sharma books are the contribution of experts in the particular field. Therefore, these solutions are also verified, and the students need not worry about the accuracy.

4. How to download the RD Sharma solution books from the Career 360 website?

Visit the Career 360 website and type the subject's name and grade for the solution book you require. The search query will display the book; click on the download button to save the RD Sharma book on your device.