First slide

Theorems of probability

Question

Consider a set P containing n elements. A subset A of P is drawn and there after set P is reconstructed. Now one
more subset B of P is drawn. Probability of drawing sets A and B so that $A \cap B$ has exactly one element is

Moderate

A

$(3 / 4)^{n} \cdot n$
B

$n \cdot (3 / 4)^{n - 1}$
C

$(n - 1) \cdot (3 / 4)^{n}$
D

None of these

Solution

$Let x_{i} be any element of set P, we have following possibilities$
$\begin{aligned} (1) x_{i} \in A, x_{i} \in B \\ (2) x_{i} \in A, x_{i} \notin B \\ (3) x_{i} \notin A, x_{i} \in B \\ (4) x_{i} \notin A, x_{i} \notin B \end{aligned}$
Clearly, the element A o B it it belongs to A and B both.
Thus out of these 4 ways only first way is favorable.
Now the element that we want to be in the intersection can be chosen
in'n' different ways. Hence required probability is $n \cdot (3 / 4)^{n - 1}$

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