JEE MAIN - Mathematics (2021 - 20th July Evening Shift - No. 7)
Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1 $$-$$ k), the probability that exactly one of B and C occurs is (1 $$-$$ 2k), the probability that exactly one of C and A occurs is (1 $$-$$ k) and the probability of all A, B and C occur simultaneously is k2, where 0 < k < 1. Then the probability that at least one of A, B and C occur is :
greater than $${1 \over 8}$$ but less than $${1 \over 4}$$
greater than $${1 \over 2}$$
greater than $${1 \over 4}$$ but less than $${1 \over 2}$$
exactly equal to $${1 \over 2}$$
Explanation
$$P(\overline A \cap B) + P(A \cap \overline B ) = 1 - k$$
$$P(\overline A \cap C) + P(A \cap \overline C ) = 1 - 2k$$
$$P(\overline B \cap C) + P(B \cap \overline C ) = 1 - k$$
$$P(A \cap B \cap C) = {k^2}$$
$$P(A) + P(B) - 2P(A \cap B) = 1 - k$$ .....(i)
$$P(B) + P(C) - 2P(B \cap C) = 1 - k$$ ..... (ii)
$$P(C) + P(A) - 2P(A \cap C) = 1 - 2k$$ ..... (iii)
$$(i) + (ii) + (iii)$$
$$P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) = {{ - 4k + 3} \over 2}$$
So,
$$P(A \cup B \cup C) = {{ - 4k + 3} \over 2} + {k^2}$$
$$P(A \cup B \cup C) = {{2{k^2} - 4k + 3} \over 2}$$
$$ = {{2{{(k - 1)}^2} + 1} \over 2}$$
$$P(A \cup B \cup C) > {1 \over 2}$$
$$P(\overline A \cap C) + P(A \cap \overline C ) = 1 - 2k$$
$$P(\overline B \cap C) + P(B \cap \overline C ) = 1 - k$$
$$P(A \cap B \cap C) = {k^2}$$
$$P(A) + P(B) - 2P(A \cap B) = 1 - k$$ .....(i)
$$P(B) + P(C) - 2P(B \cap C) = 1 - k$$ ..... (ii)
$$P(C) + P(A) - 2P(A \cap C) = 1 - 2k$$ ..... (iii)
$$(i) + (ii) + (iii)$$
$$P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) = {{ - 4k + 3} \over 2}$$
So,
$$P(A \cup B \cup C) = {{ - 4k + 3} \over 2} + {k^2}$$
$$P(A \cup B \cup C) = {{2{k^2} - 4k + 3} \over 2}$$
$$ = {{2{{(k - 1)}^2} + 1} \over 2}$$
$$P(A \cup B \cup C) > {1 \over 2}$$
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