Let E and F be two independent events. The pr | vedclass

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(B) Let $P(E) = x$ and $P(F) = y$. Since $E$ and $F$ are independent,$P(E \cap F) = xy$.The probability that exactly one of them occurs is $P(E)P(F')

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$(A, D)$

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(B) Let $P(E) = x$ and $P(F) = y$. Since $E$ and $F$ are independent,$P(E \cap F) = xy$.The probability that exactly one of them occurs is $P(E)P(F') + P(F)P(E') = x(1-y) + y(1-x) = x - xy + y - xy = x + y - 2xy = \frac{11}{25} \quad \dots (1)$.The probability that none of them occurs is $P(E' \cap F') = P(E')P(F') = (1-x)(1-y) = 1 - x - y + xy = \frac{2}{25} \quad \dots (2)$.From $(2)$,$1 - (x+y) + xy = \frac{2}{25} \Rightarrow x+y - xy = 1 - \frac{2}{25} = \frac{23}{25}$.Let $S = x+y$ and $P = xy$. Then $(1)$ becomes $S - 2P = \frac{11}{25}$ and $(2)$ becomes $S - P = \frac{23}{25}$.Subtracting the equations: $(S - P) - (S - 2P) = \frac{23}{25} - \frac{11}{25} \Rightarrow P = \frac{12}{25}$.Substituting $P$ into $S - P = \frac{23}{25}$,we get $S = \frac{23}{25} + \frac{12}{25} = \frac{35}{25} = \frac{7}{5}$.Now,$x$ and $y$ are roots of the quadratic equation $t^2 - St + P = 0$,i.e.,$t^2 - \frac{7}{5}t + \frac{12}{25} = 0$.$25t^2 - 35t + 12 = 0 \Rightarrow 25t^2 - 15t - 20t + 12 = 0 \Rightarrow 5t(5t-3) - 4(5t-3) = 0$.$(5t-4)(5t-3) = 0$,so $t = \frac{4}{5}$ or $t = \frac{3}{5}$.Thus,${P(E), P(F)} = \{\frac{3}{5}, \frac{4}{5}\}$. This corresponds to options $(A)$ and $(D)$.

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