(a) The probability of hitting the target by a trained sniper is three times the probability of not hitting th

Question

Q. (a) The probability of hitting the target by a trained sniper is three times the probability of not hitting the target on a stormy day due to high wind speed. The sniper fired two shots on the target on a stormy day when wind speed was very high. Find the probability that

(i) target is hit

(ii) atleast one shot misses the target.

(OR)

(b) Mother, Father and Son line up at random for a family picture. Let events E: Son on one end and F: Father in the middle. Find

P(E/F)

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★

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Part (a)Concept understanding — Event Independence

Event Independence

Two events are independent when the occurrence of one does not change the probability of the other. Toss a coin and roll a die: the coin landing heads tells you nothing about whether the die shows a six. Contrast this with drawing cards without replacement, where the first draw does change the odds for the second — those events are dependent.

From Conditional Probability to a Clean Test

"Knowing $B$ doesn't change $A$ " means $P(A \mid B) = P(A)$ . Substituting the definition $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ and clearing the fraction gives the symmetric form used in practice:

P(A \cap B) = P(A)\,P(B).

Events $A$ and $B$ are independent exactly when the probability of both occurring equals the product of their individual probabilities. This version is preferred because it needs no non-zero condition and treats $A$ and $B$ alike.

A Quick Check

Roll a fair die. Let $A = \{2,4,6\}$ (even) and $B = \{4,5,6\}$ (greater than 3). Then $P(A) = P(B) = \tfrac{1}{2}$ , and $A \cap B = \{4,6\}$ so $P(A \cap B) = \tfrac{1}{3}$ . Since $\tfrac{1}{3} \neq \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4}$ , these events are not independent.

Three or More Events

Events $A, B, C$ are mutually independent only if all four conditions hold: the three pairwise products and

P(A \cap B \cap C) = P(A)\,P(B)\,P(C).

Pairwise independence alone is not enough to guarantee mutual independence. …

Part (b)Concept understanding — Conditional Probability

Conditional Probability

Roll a die and ask "what is the chance of an even number?" — that is $3/6$ . But suppose someone tells you the result is greater than 3. Now you are no longer looking at all six faces, only at $\{4,5,6\}$ , and two of those (4 and 6) are even, so the probability becomes $2/3$ . That change — from the probability of $A$ to the probability of $A$ given that $B$ has already occurred — is conditional probability.

The Idea: Shrink the Sample Space

Conditioning on $B$ throws away every outcome where $B$ is false and treats $B$ as the new "whole world." You measure $A$ only against what is still possible.

Note

Think of filtering a table of data: unconditional probability uses every row; conditional probability keeps only the rows where the condition is true.

The Definition

For events $A$ and $B$ with $P(B) > 0$ ,

P(A \mid B) = \frac{P(A \cap B)}{P(B)}.

We divide by $P(B)$ to rescale so that $B$ itself has probability $1$ ; the surviving part of $A$ is the overlap $A \cap B$ . Checking the die: $P(A \cap B) = P(\{4,6\}) = \tfrac{2}{6}$ and $P(B) = \tfrac{3}{6}$ , so $P(A \mid B) = \dfrac{2/6}{3/6} = \dfrac{2}{3}$ , matching the intuition.

Tip

Rearranging gives the multiplication rule $P(A \cap B) = P(A \mid B)\,P(B)$ , which is usually the easier way to compute a joint probability when a problem says "given that."

Two Cautions

$P(A \mid B)$ and $P(B \mid A)$ are generally not equal; swapping them is the classic mistake. They are linked by Bayes' theorem, $P(A \mid B) = \dfrac{P(B \mid A)\,P(A)}{P(B)}$ . …

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(a) Hit prob $p=\frac34$ , miss $q=\frac14$ : (i) $P(\text{target hit})=\frac{15}{16}$ , (ii) $P(\text{at least one miss})=\frac{7}{16}$ . (b) With Father fixed in the middle the Son is forced to an end, so $P(E\mid F)=1$ .

Part (a) — Sniper (two shots)

Let $p=P(\text{hit})$ and $q=P(\text{miss})$ on the stormy day. We are told the hit probability is three times the miss probability, so $p=3q$ , and since a shot either hits or misses, $p+q=1$ :

3q+q=1\Rightarrow q=\tfrac14,\qquad p=\tfrac34.

The two shots are fired independently, so the outcome of one does not affect the other.

Target is hit. "The target is hit" means at least one of the two shots lands. The complement is "both shots miss": $P(\text{target hit})=1-P(\text{both miss})=1-q^2=1-\left(\tfrac14\right)^2=1-\tfrac1{16}=\frac{15}{16}.$
At least one shot misses. The complement of this is "both shots hit": …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 143 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$\dfrac{2}{3}$
›Reveal solutionSolution
The assertion is true: given the outcome is odd, the probability it is a prime is $\frac{2}{3}$ . The reason states the correct conditional probability formula. Since the reason directly justifies the calculation in the assertion, both are true and the reason is the correct explanation.

Concept first — Conditional probability asks: If we already know that event $B$ has occurred, what is the probability that event $A$ also occurs? The sample space shrinks from all possible outcomes to just those in $B$ . The formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$ is the precise way to compute this reduced probability.

Here, the die is unbiased, so each face $\{1,2,3,4,5,6\}$ has probability $\frac{1}{6}$ . The assertion involves two events:

$A$ : the number is prime. On a die, the primes are $2,3,5$ .

$B$ : the number is odd. The odd numbers are $1,3,5$ .

The condition "given that the number is odd" means we restrict attention to $B = \{1,3,5\}$ . Among these three equally likely outcomes, the primes are $3$ and $5$ — that's two out of three. So the conditional probability is $\frac{2}{3}$ .

Now let's verify step by step using the formula in Reason (R).

Define the events precisely.

$A = \{2,3,5\}$ , $B = \{1,3,5\}$ .

The sample space $S = \{1,2,3,4,5,6\}$ .

Compute $P(B)$ .

$B$ has 3 outcomes, each with probability $\frac{1}{6}$ , so $P(B) = \frac{3}{6} = \frac{1}{2}$ .

Compute $P(A \cap B)$ .

$A \cap B$ = numbers that are both prime and odd = $\{3,5\}$ . That's 2 outcomes, so $P(A \cap B) = \frac{2}{6} = \frac{1}{3}$ .

Apply the formula from Reason (R).

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/3}{1/2} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3}.$
This matches the assertion exactly. …
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CBSE 2026Set 65/3/11 markMCQ
$E$
›Reveal solutionSolution
We use the property of independent events, $P(E \cap F) = P(E)P(F)$ , along with the union formula to first find $P(F)$ . Then, we use the fact that for independent events, $P(E|F) = P(E)$ and $P(F|E) = P(F)$ , to calculate the required difference. The final result is $\boxed{\frac{1}{70}}$ .

The core of this problem lies in understanding how the concept of "independent events" simplifies probability calculations, especially when dealing with unions and conditional probabilities.

When two events, $E$ and $F$ , are independent, it means that the occurrence of one event does not affect the probability of the other event occurring. This has two crucial implications:

Intersection Probability: The probability of both $E$ and $F$ happening, $P(E \cap F)$ , is simply the product of their individual probabilities: $P(E \cap F) = P(E)P(F)$ .

Conditional Probability: The probability of $E$ happening given that $F$ has already happened, $P(E|F)$ , is just the probability of $E$ , because $F$ 's occurrence doesn't change $E$ 's likelihood. So, $P(E|F) = P(E)$ . Similarly, $P(F|E) = P(F)$ .

We are given $P(E)$ , $P(E \cup F)$ , and that $E$ and $F$ are independent. Our strategy will be to first use the formula for the union of events, combined with the independence property, to find $P(F)$ . Once we have $P(F)$ , we can directly use the independence property to find $P(E|F)$ and $P(F|E)$ , and then calculate their difference.

Find $P(F)$ using the union formula and independence.

The general formula for the probability of the union of two events is:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
Since $E$ and $F$ are independent, we can substitute $P(E \cap F)$ with $P(E)P(F)$ :

$P(E \cup F) = P(E) + P(F) - P(E)P(F)$

Now, substitute the given values: $P(E) = \frac{3}{10}$ and $P(E \cup F) = \frac{1}{2}$ .

$\frac{1}{2} = \frac{3}{10} + P(F) - \frac{3}{10} P(F)$

To solve for $P(F)$ , group the terms involving $P(F)$ :

$\frac{1}{2} = \frac{3}{10} + P(F) \left(1 - \frac{3}{10}\right)$

$\frac{1}{2} = \frac{3}{10} + P(F) \left(\frac{7}{10}\right)$

Subtract $\frac{3}{10}$ from both sides:

$\frac{1}{2} - \frac{3}{10} = P(F) \left(\frac{7}{10}\right)$

To subtract the fractions on the left, find a common denominator, which is $10$ :

$\frac{5}{10} - \frac{3}{10} = P(F) \left(\frac{7}{10}\right)$

$\frac{2}{10} = P(F) \left(\frac{7}{10}\right)$

Now, isolate $P(F)$ by multiplying both sides by $\frac{10}{7}$ : …
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CBSE 2026Set V11 markMCQ
$Q. The probability of obtaining an even prime number on each die when a pair of dice is rolled$
(a) $\frac{1}{36}$
(b) $\frac{1}{6}$
(c) $\frac{1}{18}$
(d) $\frac{1}{4}$
›Reveal solutionSolution

The even prime is $2$ ; $P(2\text{ on each of two dice})=\tfrac{1}{6}\cdot\tfrac{1}{6}=\tfrac{1}{36}$ ; answer (a).

The only even prime number is $2$ . For one die, $P(\text{show }2)=\dfrac{1}{6}$ . The two dice are independent, so …
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CBSE 2026Set V11 markMCQ
$A$
(a) $1.2$
(b) $0.12$
(c) $0.7$
(d) $\frac{3}{4}$
›Reveal solutionSolution

Independence gives $P(A\cap B)=P(A)P(B)=0.12$ ; answer (b).

For independent events $A$ and $B$ ,

$P(A\cap B)=P(A)\cdot P(B)=0.3\times0.4=0.12.$ …
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CBSE 2026Set V11 mark
$[0, 3, -1, 2, -2, 1]$
›Reveal solutionSolution

Since $S\cap F=F$ , the conditional probability $P(S\mid F)=1$ .

By the definition of conditional probability (with $P(F)\neq0$ ),

$P(S\mid F)=\frac{P(S\cap F)}{P(F)}.$ …
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CBSE 2026Set CX1 markMCQ
$3P(A)=P(B)=\dfrac{5}{13}$
(a) $\dfrac{20}{39}$
(b) $\dfrac{16}{39}$
(c) $\dfrac{11}{39}$
(d) $\dfrac{14}{39}$
›Reveal solutionSolution

Using $P(A\cap B)=P(A/B)P(B)$ and the addition rule gives $P(A\cup B)=\dfrac{14}{39}$ — option (d).

Given: $3P(A)=P(B)=\dfrac{5}{13}$ and $P(A/B)=\dfrac{2}{5}$ .

So $P(B)=\dfrac{5}{13}$ and $P(A)=\dfrac{1}{3}\cdot\dfrac{5}{13}=\dfrac{5}{39}$ .

Intersection (multiplication rule):

$P(A\cap B)=P(A/B)\,P(B)=\dfrac{2}{5}\cdot\dfrac{5}{13}=\dfrac{2}{13}.$

…

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CBSE 2026Set A1 markMCQ
$A$
(a) $P(A) + P(B) + P(C)$
(b) $P(A) - P(B) - P(C)$
(c) $P(A) \cdot P(B) \cdot P(C)$
(d) None of these
›Reveal solutionSolution

For independent events, $P(A\cap B\cap C) = P(A)P(B)P(C)$ .

By definition, events $A$ , $B$ , $C$ are (mutually) independent when the probability of their joint occurrence equals the product of their individual probabilities:

$P(ABC) = P(A)\cdot P(B)\cdot P(C).$

…

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CBSE 2026Set A1 markMCQ
$P(A) = \frac{7}{13},\ P(B) = \frac{9}{13},\ P(A \cap B) = \frac{4}{13} \Rightarrow P(A/B) =$
(a) $\frac{4}{9}$
(b) $\frac{4}{7}$
(c) $\frac{12}{13}$
(d) $\frac{1}{6}$
›Reveal solutionSolution

$P(A|B) = \frac{4}{9}$ .

Use the conditional-probability definition:

$P(A|B) = \frac{P(A\cap B)}{P(B)}.$

Substitute the given values: …

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CBSE 2026Set ANNUAL1 markMCQ
$P(B)=0.5$
(a) $\frac{15}{23}$
(b) $\frac{16}{25}$
(c) $\frac{16}{27}$
(d) $\frac{16}{23}$
›Reveal solutionSolution

By the definition of conditional probability, $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{16}{25}$ .

The conditional probability of $A$ given $B$ is defined as

$P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B) eq0$

…

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CBSE 2026Set ANNUAL1 markMCQ
$P(A) = 0.3$
(a) $0.58$
(b) $0.70$
(c) $0.12$
(d) $0.10$
›Reveal solutionSolution

For independent events, $P(A\cap B)=P(A)P(B)$ , and the addition rule gives $P(A\cup B)$ .

$P(A\cap B) = 0.3\times0.4=0.12$ (independence).

…

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CBSE 2026Set ANNUAL1 mark
$Q. A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?$
›Reveal solutionSolution

List the equally likely outcomes for two children, restrict to those with at least one boy, then find the fraction that are both boys.

Sample space $=\{BB, BG, GB, GG\}$ , each equally likely.

Given at least one boy: reduced sample space $=\{BB,BG,GB\}$ (3 outcomes).

…

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CBSE 2026Set ANNUAL1 markMCQ
$P(A) = 0.8$
(a) 0.8
(b) 0.5
(c) 0.32
(d) 0.4
›Reveal solutionSolution

Use the multiplication rule of conditional probability: $P(A\cap B) = P(B\mid A)\cdot P(A)$ .

Given $P(A)=0.8$ , $P(B\mid A)=0.4$ .

$P(A\cap B) = P(B\mid A)\cdot P(A) = 0.4 \times 0.8 = 0.32$ .

…

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Q.(a) The probability of hitting the target by a trained sniper is three times the probability of not hitting the target on a stormy day due to high wind speed. The sniper fired two shots on the target on a stormy day when wind speed was very high. Find the probability that

Part (a) — Sniper (two shots)

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