For the three events $A,B\ and\ C,\ P$ (exactly one of the events $A\ or\ B$ occurs)= $P$(exactly one of the 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

Question

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

Dr. Abubacker · Accepted Answer

(a)
We know that, $P$ (exactly one of $A\ or\ B$ occurs)
$= P(A) + P(B) - 2P(A \cap B)$
Therefore, $P(A) + P(B) - 2P(A \cap B) = p$ …(i)
Similarly, $P(B) + P(C) - 2P(B \cap C) = p$ ...(ii)
and $P(C) + P(A) - 2P(C \cap A) = p$ ….(iii)
Adding Eqs. (i), (ii) and (iii), we get
$2\left\lbrack P(A) + P(B) + P(C) - \ P(A \cap B) - P(B \cap C) - P(C \cap A) \right\rbrack = 3P$
$\Rightarrow P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) = \frac{3P}{2}$ ….(iv)
We are also given that
$P(A \cap B \cap C) = p^{2}$ …(v)
Now, $P$ (at least one of $A,B\ and\ C)$
$=$ $P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$
[from Eqs. (iv) and (v)]
$= \frac{3p}{2} + p^{2}$
$= \frac{3p + p^{2}}{2}$

Explanation

Reviewed by Dr. Abubacker

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