Anand Classes provides comprehensive and exam-focused NCERT Solutions for Probability Miscellaneous Exercise of Class 12 Chapter 13, helping students master complex concepts with clarity and confidence. These notes are designed to simplify problem-solving, strengthen conceptual understanding, and support effective board exam preparation. With step-by-step solutions and easy-to-download study material, students can enhance their practice and score higher in exams. Click the print button to download study material and notes.
NCERT Question.1 : A and B are two events such that $P(A) \neq 0$. Find $P(B \mid A)$, if:
(i) $A$ is a subset of $B$
(ii) $A \cap B = \emptyset$
Solution
We know that the conditional probability is defined as:
$$
P(B \mid A) = \frac{P(A \cap B)}{P(A)}
$$
(i) When $A \subset B$
If $A$ is a subset of $B$, then
$$
A \cap B = A
$$
$$
P(A \cap B) = P(A)
$$
Thus,
$$
P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)}{P(A)} = 1
$$
(ii) When $A \cap B = \emptyset$
If $A$ and $B$ are mutually exclusive, then
$$
P(A \cap B) = 0
$$
Thus,
$$
P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0}{P(A)} = 0
$$
Final Result
- (i) $P(B \mid A) = 1$
- (ii) $P(B \mid A) = 0$
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NCERT Question.2 : A couple has two children,
(i) Find the probability that both children are males, if it is known that at least one of the children is male.
(ii) Find the probability that both children are females, if it is known that the elder child is a female.
Solution
The sample space for two children is:
$$
S = \{(b, b), (b, g), (g, b), (g, g)\}
$$
where $b$ = boy and $g$ = girl.
(i) Probability that both are males given at least one male
Let $A$ = event that both children are males
$$
A = \{(b, b)\}
$$
Let $B$ = event that at least one child is male
$$
B = \{(b, b), (b, g), (g, b)\}
$$
Then,
$$
(A \cap B) = \{(b, b)\}
$$
$$
P(A \cap B) = \frac{1}{4}, \quad P(B) = \frac{3}{4}
$$
Hence, conditional probability:
$$
P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{3/4} = \frac{1}{3}
$$
(ii) Probability that both are females given elder is female
Let $C$ = event that both children are females
$$
C = \{(g, g)\}
$$
Let $D$ = event that elder child is female
$$
D = \{(g, g), (g, b)\}
$$
Then,
$$
(C \cap D) = \{(g, g)\}
$$
$$
P(C \cap D) = \frac{1}{4}, \quad P(D) = \frac{2}{4} = \frac{1}{2}
$$
Hence, conditional probability:
$$
P(C \mid D) = \frac{P(C \cap D)}{P(D)} = \frac{1/4}{1/2} = \frac{1}{2}
$$
Final Result
- (i) $P(\text{both male} \mid \text{at least one male}) = 1/3$
- (ii) $P(\text{both female} \mid \text{elder female}) = 1/2$
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NCERT Question.3 : Suppose that 5% of men and 0.25% of women have grey hair. A grey-haired person is selected at random. What is the probability of this person being male? Assume that there is an equal number of males and females.
Solution
Let $M$ = event that the person selected is male
$F$ = event that the person selected is female
$G$ = event that the person selected has grey hair
We need to find $P(M \mid G)$. Using Bayes’ theorem:
$$
P(M \mid G) = \frac{P(M) \cdot P(G \mid M)}{P(M) \cdot P(G \mid M) + P(F) \cdot P(G \mid F)}
$$
Given:
$$
P(G \mid M) = 0.05, \quad P(G \mid F) = 0.0025
$$
Since there is an equal number of males and females:
$$
P(M) = P(F) = \frac{1}{2}
$$
Now, substituting the values:
$$
P(M \mid G) = \frac{\frac{1}{2} \cdot 0.05}{\frac{1}{2} \cdot 0.05 + \frac{1}{2} \cdot 0.0025} $$
$$P(M \mid G) = \frac{0.025}{0.025 + 0.00125} = \frac{0.025}{0.02625} = \frac{100}{105} = \frac{20}{21}
$$
Final Result
$$
P(\text{person is male} \mid \text{grey hair}) = \frac{20}{21}
$$
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NCERT Question.4 : Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?
Solution
Let $p$ denote the probability of a person being right-handed and $q$ denote the probability of being left-handed.
Given:
$$
p = 0.9, \quad q = 1 – p = 0.1
$$
Let $X$ be the number of right-handed people in a sample of 10. Then $X$ follows a binomial distribution:
$$
X \sim \text{Binomial}(n=10, p=0.9)
$$
We are asked to find the probability that at most 6 are right-handed:
$$
P(X \le 6) = 1 – P(X > 6) = 1 – P(X = 7,8,9,10)
$$
Using the binomial formula:
$$
P(X = r) = {^{10}C_r} p^r q^{10-r}
$$
Thus,
$$
P(X > 6) = \sum_{r=7}^{10} {^{10}C_r} (0.9)^r (0.1)^{10-r}
$$
Finally,
$$
P(X \le 6) = 1 – \sum_{r=7}^{10} {^{10}C_r} (0.9)^r (0.1)^{10-r}
$$
Final Result
$$
P(\text{at most 6 right-handed}) = 1 – \sum_{r=7}^{10} {^{10}C_r} (0.9)^r (0.1)^{10-r}
$$
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NCERT Question.5 : If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?
Solution
A leap year has a total of 366 days, which is 52 weeks and 2 extra days.
In 52 weeks, there are exactly 52 Tuesdays. The extra 2 days determine whether there will be an additional Tuesday, making it 53.
The possible combinations of the 2 extra days are:
$$
\{(\text{Monday, Tuesday}), (\text{Tuesday, Wednesday}), \\[1em](\text{Wednesday, Thursday}), (\text{Thursday, Friday}), \\[1em](\text{Friday, Saturday}), (\text{Saturday, Sunday}), (\text{Sunday, Monday})\}
$$
There are 7 equally likely cases for the extra 2 days. Out of these, Tuesday appears in 2 cases:
$$
(\text{Monday, Tuesday}) \text{ and } (\text{Tuesday, Wednesday})
$$
Hence, the probability that a leap year has 53 Tuesdays is:
$$
P(\text{53 Tuesdays}) = \frac{2}{7}
$$
Final Result
$$
P(\text{53 Tuesdays in a leap year}) = \frac{2}{7}
$$
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NCERT Question.6 : Suppose we have four boxes A, B, C, D containing colored marbles as given below:
| BOX | RED | WHITE | BLACK |
|---|---|---|---|
| A | 1 | 6 | 3 |
| B | 6 | 2 | 2 |
| C | 8 | 1 | 1 |
| D | 0 | 6 | 4 |
One of the boxes is selected at random and a single marble is drawn from it. If the marble is red, find the probability that it was drawn from box A, box B, or box C.
Solution
Let $R$ be the event of drawing a red marble. Let $E_A$, $E_B$, and $E_C$ denote the events of selecting boxes A, B, and C, respectively.
The total number of marbles in all boxes:
$$
1+6+3 + 6+2+2 + 8+1+1 + 0+6+4 = 40
$$
The total number of red marbles:
$$
1+6+8+0 = 15
$$
So,
$$
P(R) = \frac{15}{40} = \frac{3}{8}
$$
Probability that the red marble is from box A
$$
P(E_A|R) = \frac{P(E_A \cap R)}{P(R)} = \frac{1/40}{3/8} = \frac{1}{15}
$$
Probability that the red marble is from box B
$$
P(E_B|R) = \frac{P(E_B \cap R)}{P(R)} = \frac{6/40}{3/8} = \frac{2}{5}
$$
Probability that the red marble is from box C
$$
P(E_C|R) = \frac{P(E_C \cap R)}{P(R)} = \frac{8/40}{3/8} = \frac{8}{15}
$$
Final Result
$$
P(\text{Red from A}) = \frac{1}{15}, \quad P(\text{Red from B}) = \frac{2}{5}, \quad P(\text{Red from C}) = \frac{8}{15}
$$
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NCERT Question.7 : Assume that the chances of a patient having a heart attack are 40%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by 30%, and a prescription of a certain drug reduces its chances by 25%. At a time, a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.
Solution
Let $X$ denote the event that the patient suffers a heart attack. Let $A_1$ denote the event that the patient follows a course of meditation and yoga, and $A_2$ denote the event that the patient takes the drug prescription.
Given:
$$
P(X) = 0.40, \quad P(A_1) = P(A_2) = \frac{1}{2}
$$
The probability of a heart attack given the patient followed meditation and yoga:
$$
P(X|A_1) = 0.40 \times (1 – 0.30) = 0.40 \times 0.70 = 0.28
$$
The probability of a heart attack given the patient took the drug prescription:
$$
P(X|A_2) = 0.40 \times (1 – 0.25) = 0.40 \times 0.75 = 0.30
$$
We need to find the probability that the patient followed meditation and yoga given that they had a heart attack:
$$
P(A_1|X) = \frac{P(A_1) P(X|A_1)}{P(A_1) P(X|A_1) + P(A_2) P(X|A_2)}
$$
Substituting the values:
$$
P(A_1|X) = \frac{\frac{1}{2} \times 0.28}{\frac{1}{2} \times 0.28 + \frac{1}{2} \times 0.30}
= \frac{0.14}{0.14 + 0.15} = \frac{0.14}{0.29} = \frac{14}{29}
$$
Final Result
$$
P(\text{Patient followed meditation and yoga | Heart attack}) = \frac{14}{29}
$$
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