Anand Classes offers clear and exam-focused Probability NCERT Solutions for Exercise 13.3 Class 12 Chapter 13 (Set-1) to help students master advanced probability concepts with ease. These solutions follow the latest NCERT syllabus and provide step-by-step explanations, solved examples, and accurate methods ideal for CBSE board preparation. Students can rely on these notes for quick revision, deeper understanding, and enhanced problem-solving confidence. Click the print button to download study material and notes.
Access Probability NCERT Solutions for Exercise 13.3 Class 12 Chapter 13
NCERT Question 1 : An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Solution
When the first draw gives a red ball, two additional red balls are put in the urn so that its contents are 7 (= 5 + 2) red and 5 black balls. When the first draw gives a black ball, two additional black balls are put in the urn so that its contents are 5 red and 7 (= 5 + 2) black balls.
Let $B$: first ball drawn is black
$R$: second ball is red
Probability that the second ball is red = Probability that the first ball is black × Probability that the second ball is red if first is black + Probability that first ball is red × Probability that second ball is red if first is red.
Probability first ball is black and second ball is red
Probability first ball is black
$$\frac{5}{5+5} = \frac{5}{10} = \frac{1}{2}$$
Now, there are $5$ Red and $(5+2)=7$ Black balls in the urn.
So, Probability second ball is red when first ball is black
$$\frac{5}{5+7} = \frac{5}{12}$$
Probability first ball is red and second ball is red
Probability first ball is red
$$1 – P(\text{first is black}) = 1 – \frac{1}{2} = \frac{1}{2}$$
Now, there are $(5+2)=7$ Red and $5$ Black balls in the urn.
So, Probability second ball is red when first ball is red
$$\frac{7}{5+7} = \frac{7}{12}$$
Thus, Probability that the second ball is red
$$\frac{1}{2} \times \frac{5}{12} + \frac{1}{2} \times \frac{7}{12}$$
$$= \frac{5}{24} + \frac{7}{24} = \frac{12}{24} = \frac{1}{2}$$
Therefore, Probability that the second ball is red is
$$\frac{1}{2}$$
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NCERT Question 2 : A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Solution
Bag 1 : 4 Red (R) 4 Black (B) balls
Bag 2 :2 Red (R) 6 Black (B) balls
Let $B_{1}$ : ball is drawn from Bag I
$B_{2}$ : ball is drawn from Bag II
$R$ : ball drawn is red
We need to find Probability that the ball is drawn from Bag I, if the ball is red
$$P(B_{1}\mid R)$$
Using the formula,
$$P(B_{1}\mid R)=\frac{P(B_{1});P(R\mid B_{1})}{P(B_{1});P(R\mid B_{1}) + P(B_{2})P(R\mid B_{2})}$$
Probability that ball is drawn from Bag I
$$P(B_{1})=\frac{1}{2}$$
Probability that ball is red if drawn from Bag I
$$P(R\mid B_{1})=\frac{4}{4+4}=\frac{4}{8}=\frac{1}{2}$$
Probability that ball is drawn from Bag II
$$P(B_{2})=\frac{1}{2}$$
Probability that ball is red if drawn from Bag II
$$P(R\mid B_{2})=\frac{2}{2+6}=\frac{2}{8}=\frac{1}{4}$$
Putting values in the formula,
$$P(B_{1}\mid R)=\frac{\frac{1}{2}\times\frac{1}{2}}{\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{4}}$$
$$=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}=\frac{\frac{2}{8}}{\frac{3}{8}}=\frac{2}{3}$$
Therefore, the required probability is
$$\frac{2}{3}$$
Enhance your understanding of conditional probability with expertly crafted notes and solutions from Anand Classes, ideal for CBSE and competitive exams.
NCERT Question 3 : Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostler?
Solution
Let $H$ : student selected is a hostler
$D$ : student selected is a day scholar
$A$ : student has an A grade
We need to find
$$P(H\mid A)$$
Using the formula,
$$P(H\mid A)=\frac{P(H)P(A\mid H)}{P(D)P(A\mid D)+P(H)P(A\mid H)}$$
Probability student is a hostler
$$P(H)=\frac{60}{100}=0.6$$
Probability of A grade if hostler
$$P(A\mid H)=\frac{30}{100}=0.3$$
Probability student is a day scholar
$$P(D)=\frac{40}{100}=0.4$$
Probability of A grade if day scholar
$$P(A\mid D)=\frac{20}{100}=0.2$$
Substituting values,
$$P(H\mid A)=\frac{0.6\times 0.3}{0.4\times 0.2 + 0.6\times 0.3}$$
$$=\frac{0.18}{0.08+0.18}=\frac{0.18}{0.26}=\frac{18}{26}=\frac{9}{13}$$
Therefore, the required probability is
$$\frac{9}{13}$$
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NCERT Question 4: In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$. What is the probability that the student knows the answer given that he answered it correctly?
Solution
Let $A$: student knows the answer
$B$: student guesses
$C$: student answers correctly
We need to find $P(A|C)$.
$$P(A|C)=\frac{P(A)P(C|A)}{P(A)P(C|A)+P(B)P(C|B)}$$
Given
$P(A)=\frac{3}{4}$
$P(C|A)=1$
$P(B)=\frac{1}{4}$
$P(C|B)=\frac{1}{4}$
Substituting the values:
$$P(A|C)=\frac{\frac{3}{4} \times 1}{\frac{1}{4}\times\frac{1}{4}+\frac{3}{4}\times 1}$$
$$P(A|C)=\frac{\frac{3}{4}}{\frac{1}{16}+\frac{3}{4}}$$
$$P(A|C)=\frac{\frac{3}{4}}{\frac{13}{16}}$$
$$P(A|C)=\frac{12}{13}$$
Final Result
$$\boxed{\frac{12}{13}}$$
This chapter strengthens concepts of conditional probability and Bayes’ theorem, essential for JEE and CBSE Class 12 exams. For more such high-quality explanations and practice material, explore advanced notes and question banks by Anand Classes.
NCERT Question 5: A laboratory blood test is 99% effective in detecting a certain disease when it is in fact present. However, the test also yields a false positive result for 0.5% of healthy persons tested. If 0.1% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Solution
Let
$A$: person has the disease
$B$: person does not have the disease
$C$: test result is positive
We need to find $P(A|C)$.
Given
$P(A)=0.001$
$P(C|A)=0.99$
$P(B)=0.999$
$P(C|B)=0.005$
Using Bayes’ theorem:
$$P(A|C)=\frac{P(A)P(C|A)}{P(B)P(C|B)+P(A)P(C|A)}$$
Substituting:
$$P(A|C)=\frac{0.001\times 0.99}{0.999\times 0.005+0.001\times 0.99}$$
$$P(A|C)=\frac{0.00099}{0.004995+0.00099}$$
$$P(A|C)=\frac{0.00099}{0.005985}$$
$$P(A|C)=\frac{99}{598.5}=\frac{990}{5985}=\frac{22}{133}$$
Final Result
$$\boxed{\frac{22}{133}}$$
Understanding such real-life applications of Bayes’ theorem is crucial for mastering probability in Class 12 and competitive exams. For more expertly crafted explanations and notes, explore the comprehensive study resources by Anand Classes.
NCERT Question 6: There are three coins. One is a two-headed coin, another is a biased coin that shows heads 75% of the time, and the third is an unbiased coin. One of the three coins is chosen at random and tossed. It shows heads. What is the probability that it was the two-headed coin?
Solution
Let
$C_1$: two-headed coin
$C_2$: biased coin
$C_3$: unbiased coin
$H$: head appears
We need to find $P(C_1|H)$.
Given:
$P(C_1)=\frac13$, $P(H|C_1)=1$
$P(C_2)=\frac13$, $P(H|C_2)=\frac34$
$P(C_3)=\frac13$, $P(H|C_3)=\frac12$
Using Bayes’ theorem:
$$P(C_1|H)=\frac{P(C_1)P(H|C_1)}{P(C_1)P(H|C_1)+P(C_2)P(H|C_2)+P(C_3)P(H|C_3)}$$
Substituting:
$$P(C_1|H)=\frac{\frac13\times 1}{\frac13\times 1+\frac13\times \frac34+\frac13\times \frac12}$$
Factor out $\frac13$:
$$P(C_1|H)=\frac{1}{1+\frac34+\frac12}$$
$$P(C_1|H)=\frac{1}{\frac94}=\frac49$$
Final Result
$$\boxed{\frac{4}{9}}$$
Mastering Bayes’ theorem with such examples strengthens your probability concepts for Class 12 and competitive exams. For more high-quality explanations and revision notes, explore the comprehensive study material by Anand Classes.
NCERT Question 7: An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident are $0.01$, $0.03$ and $0.15$ respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Solution
Let
$S$: scooter driver
$C$: car driver
$T$: truck driver
$A$: accident occurs
We need to find $P(S|A)$.
Given:
Total insured persons $=2000+4000+6000=12000$
$$P(S)=0.01 \qquad P(A|S)=\frac{2000}{12000}=\frac16$$
$$P(C)=0.03 \qquad P(A|C)=\frac{4000}{12000}=\frac13$$
$$P(T)=0.15 \qquad P(A|T)=\frac{6000}{12000}=\frac12$$
Using Bayes’ theorem:
$$P(S|A)=\frac{P(S)P(A|S)}{P(S)P(A|S)+P(C)P(A|C)+P(T)P(A|T)}$$
Substituting:
$$P(S|A)=\frac{0.01\times\frac16}{0.01\times\frac16+0.03\times\frac13+0.15\times\frac12}$$
Factor out $0.01$:
$$P(S|A)=\frac{\frac16}{\frac16+3\times\frac13+15\times\frac12}$$
Compute:
$$P(S|A)=\frac{\frac16}{\frac16+1+\frac{15}{2}}$$
Convert to common denominator:
$$P(S|A)=\frac{\frac16}{\frac{52}{6}}=\frac{1}{52}$$
Final Result
$$\boxed{\frac{1}{52}}$$
Strengthen your understanding of Bayes’ theorem and conditional probability with more practice questions available in the detailed study material by Anand Classes, ideal for Class 12 board exams and competitive exam preparation.
