CBSE 2026 · Region 2 · Set 2 · Q29 · 3 marks
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.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).
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.
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).
Marking-scheme solution
(a)
Let probability of not hitting $\displaystyle =p$
$\displaystyle \therefore$ Probability of hitting $\displaystyle =3p$
$\displaystyle p+3p=1 \Rightarrow p=\dfrac{1}{4}$
So, Probability of hitting, $\displaystyle P(H)=\dfrac{3}{4}$, Probability of not hitting, $\displaystyle P(\bar{H})=\dfrac{1}{4}$
Now, $\displaystyle P(\text{Target is hit})=1-P(\bar{H}\bar{H})=1-\dfrac{1}{4} \times \dfrac{1}{4}=\dfrac{15}{16}$
(ii)
$\displaystyle P$ (at least one shot misses the target)
$\displaystyle =1-P(HH)=1-\dfrac{3}{4} \times \dfrac{3}{4}=\dfrac{7}{16}$
There are $\displaystyle 3$ people - Mother $\displaystyle (M)$, Father $\displaystyle (F)$, Son $\displaystyle (S)$.
Sample Space $\displaystyle =\{MFS, MSF, FMS, FSM, SMF, SFM\}$
Total possible arrangements $\displaystyle =6$
Given $\displaystyle E$ : Son on one end, $\displaystyle F$ : Father in the middle
$\displaystyle P(E \cap F)=P(\text{Son on one end and Father in the middle})=\dfrac{2}{6}, P(F)=\dfrac{2}{6}$
$\displaystyle \therefore P(E \mid F)=\dfrac{P(E \cap F)}{P(F)}=1$
Alternative Method :
For $\displaystyle P(E \mid F)$, as $\displaystyle F$ has already occured which means that the father is in the middle. So, son has to be on one end.
$\displaystyle \therefore P(E \mid F)=1$ (sure event)
ProbabilityConditional ProbabilityApplyshort_answermedium
More from Probability
- Assertion: Two coins are tossed simultaneously. The probability of getting two heads, if it is known that at…2023 · asked 3×
- A shop selling electronic items sells smartphones of only three reputed companies A, B and C because chances…2025 · asked 3×
- Smoking increases the risk of lung problems. A study revealed that 170 in 1000 males who smoke develop lung…2026 · asked 3×
- The probability distribution of a random variable X is: where k is some unknown constant. The probability…2024 · asked 3×
- Recent studies suggest that roughly 12 % of the world population is left handed. Depending upon the parents,…2023 · asked 3×
- Two balls are drawn at random from a bag containing 2 red balls and 3 blue balls, without replacement. Let…2022 · asked 3×
- Let X be a random variable which assumes values x 1, x 2, x 3, x 4 such that 2 P(X=x 1)=3 P(X=x 2)=P(X=x 3)=5…2022 · asked 3×
- The probability distribution of a random variable X is given below: (i) Find the value of k. (ii) Find P(1 ≤…2023 · asked 3×
CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.