Let E and F be two independent events. The probability that exactly one of them occurs is \frac{11}{25} and the probability of none of them occurring is \frac{2}{25}. If P(T) denotes the probability of occurrence of the event T, then

Engineering

Mathematics

Addition Theorem on Probability

Question

Let E and F be two independent events. The probability that exactly one of them occurs is $\frac{11}{25}$ and the probability of none of them occurring is $\frac{2}{25}$ . If P(T) denotes the probability of occurrence of the event T, then

$P (E) = \frac{3}{5}, P (F) = \frac{4}{5}$

$P (E) = \frac{1}{5}, P (F) = \frac{2}{5}$

$P (E) = \frac{4}{5}, P (F) = \frac{3}{5}$

$P (E) = \frac{2}{5}, P (F) = \frac{1}{5}$

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P(E) (1 – P(F)) + (1 – P(E)) P(E) $= \frac{11}{25}$

P(E) + P(F) – 2P (E) P(F) = $\frac{11}{25}$ ……….(1)

(1 – P(E) – P(F)) P(E) P(F) $= \frac{2}{25}$

1 – P(E) – P(F) + P(E) P(F) $= \frac{2}{25}$ ………(2)

P(E) + P(E) – P(E) (PF) P(F) $= \frac{23}{25}$

From (1) & (2)

P(E) P(F) $= \frac{12}{25}$

And P(E) + P(F) $= \frac{7}{5}$

So either

$P (E) = \frac{4}{5}, P (F) = \frac{3}{5} and P (E) = \frac{3}{5}, P (E) = \frac{4}{5}$

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