CBSE 2022 · Region 4 · Set 3 · Q14 · 4 marks
In a game of Archery, each ring of the Archery target is valued. The centremost ring is worth $\displaystyle 10$ points and rest of the rings are allotted points $\displaystyle 9$ to $\displaystyle 1$ in sequential order moving outwards. Archer A is likely to earn $\displaystyle 10$ points with a probability of $\displaystyle 0 \cdot 8$ and Archer B is likely the earn $\displaystyle 10$ points with a probability of $\displaystyle 0 \cdot 9$.
Based on the above information, answer the following questions : If both of them hit the Archery target, then find the probability that(a)exactly one of them earns $\displaystyle 10$ points.(b)both of them earn $\displaystyle 10$ points.
In a game of Archery, each ring of the Archery target is valued. The centremost ring is worth $\displaystyle 10$ points and rest of the rings are allotted points $\displaystyle 9$ to $\displaystyle 1$ in sequential order moving outwards. Archer A is likely to earn $\displaystyle 10$ points with a probability of $\displaystyle 0 \cdot 8$ and Archer B is likely the earn $\displaystyle 10$ points with a probability of $\displaystyle 0 \cdot 9$.
Based on the above information, answer the following questions : If both of them hit the Archery target, then find the probability that
(a)
exactly one of them earns $\displaystyle 10$ points.
(b)
both of them earn $\displaystyle 10$ points.
Marking-scheme solution
$\displaystyle P(A) = 0\cdot8$ , $\displaystyle P(B) = 0\cdot9$
(a)
$\displaystyle P$ (exactly one of them earns $\displaystyle 10$ points) $\displaystyle = P(A)P(\bar{B}) + P(\bar{A})P(B)$
\[= 0\cdot8 \times 0\cdot1 + 0\cdot2 \times 0\cdot9\]
\[= 0\cdot26\]
(b)
$\displaystyle P(A \cap B) = P(A)P(B)$
\[= 0\cdot8 \times 0\cdot9\]
\[= 0\cdot72\]
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.