JEE Advance - Mathematics (2011 - Paper 2 Offline - No. 10)
Let $$E$$ and $$F$$ be two independent events. The probability that exactly one of them occurs is $$\,{{11} \over {25}}$$ and the probability of none of them occurring is $$\,{{2} \over {25}}$$. If $$P(T)$$ denotes the probability of occurrence of the event $$T,$$ then
$$P\left( E \right) = {4 \over 5},P\left( F \right) = {3 \over 5}$$
$$P\left( E \right) = {1 \over 5},P\left( F \right) = {2 \over 5}$$
$$P\left( E \right) = {2 \over 5},P\left( F \right) = {1 \over 5}$$
$$P\left( E \right) = {3 \over 5},P\left( F \right) = {4 \over 5}$$
Explanation
Let $$P(E) = e$$ and $$P(F) = f$$.
$$P(E \cup F) - P(E \cap F) = {{11} \over {25}}$$
$$ \Rightarrow e + f - 2ef = {{11} \over {25}}$$ ...... (1)
$$P(\overline E \cap \overline F ) = {2 \over {25}}$$
$$ \Rightarrow (1 - e)(1 - f) = {2 \over {25}}$$
$$ \Rightarrow 1 - e - f + ef = {2 \over {25}}$$ ...... (2)
From Eqs. (1) and (2), we get
$$ef = {{12} \over {25}}$$ and $$e + f = {7 \over 5}$$
Solving, we get
$$e = {4 \over 5},\,f = {3 \over 5}$$ or $$e = {3 \over 5},\,f = {4 \over 5}$$
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