First slide

Theorems of probability

Question

$For the three events A, B and C, P (exactly one of the events A or B occurs) = P (exactly one of the two events$
$B or C occurs) = P (exactly one of the events C or A occurs) = p and P (all the three events occur simultaneously) = p^{2}$
$where 0 < p < 1 / 2 . Then the probability of at least one of the three events A, B and C occurring is$

Moderate

A

$\frac{3 p + 2 p^{2}}{2}$
B

$\frac{p + 3 p^{2}}{4}$
C

$\frac{p + 3 p^{2}}{2}$
D

$\frac{3 p + 2 p^{2}}{4}$

Solution

We know that
P(exactly one of A or B occurs) $= P (A) + P (B) - 2 P (A \cap B)$
Therefore,
$P (A) + P (B) - 2 P (A \cap B) = p - - - - - - (1)$
Similarly,
$P (B) + P (C) - 2 P (B \cap A) = p - - - - (2)$
and
$P (C) + P (A) - 2 P (C \cap A) = p - - - - - (3)$
Adding Eqs. (1), (2) and (3) we get
$\begin{array}{l} 2 [P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A)] = 3 p \\ \Rightarrow P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) = 3 p / 2 - - - (4) \end{array}$
It is also given that
$P (A \cap B \cap C) = p^{2} - - - - (5)$
Now,
P(at least one of A, B and Q)
$= P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) + P (A \cap B \cap C)$
$\begin{aligned} = \frac{3 p}{2} + p^{2} \\ = \frac{3 p + 2 p^{2}}{2} \end{aligned}$ $[Using Eqs. (4) and (5)]$

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