Question ID: #907
Let S be a set of 5 elements and $P(S)$ denote the power set of S. Let E be an event of choosing an ordered pair (A, B) from the set $P(S)\times P(S)$ such that $A \cap B=\emptyset$. If the probability of the event E is $\frac{3^{p}}{2^{q}}$ where $p, q\in\mathbb{N}$, then $p+q$ is equal to
Solution:
Number of elements in $S$, $n(S) = 5$.
Total number of subsets in $P(S) = 2^5$.
Total number of ordered pairs $(A, B)$ in $P(S) \times P(S) = 2^5 \times 2^5 = 4^5 = 2^{10}$.
For disjoint sets $A \cap B = \emptyset$, each element of $S$ has 3 choices:
1. Element is in A only.
2. Element is in B only.
3. Element is in neither A nor B.
(It cannot be in both, as intersection is empty).
So, favorable cases = $3^5$.
Probability $P(E) = \frac{3^5}{4^5} = \frac{3^5}{2^{10}}$.
Comparing with $\frac{3^p}{2^q}$:
$p = 5, q = 10$.
$$p + q = 5 + 10 = 15$$
Ans. (15)
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