Question

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn randomly one-by-one without replacement and are found to be both kings. Find the probability of the lost card being a king.

Answer 1

Answer:

$\boxed{\sf \: Required\:probability\:=\:\dfrac{1}{25} \: }$

Step-by-step explanation:

Let assume the following events

E₁ : Card lost is king.

E₂ : Card lost is not king

E : getting two cards of king.

Now,

$\sf \: P(E_1) = \dfrac{4}{52} = \dfrac{1}{13}$

$\sf \: P(E_2) = \dfrac{48}{52} = \dfrac{12}{13}$

Now,

$\sf \: P(E \: \mid \: E_1) = \dfrac{C(3,2)}{C(51,2)} = \dfrac{3 \times 2}{51 \times 50} = \dfrac{6}{2550}$

and

$\sf \: P(E \: \mid \: E_2) = \dfrac{C(4,2)}{C(51,2)} = \dfrac{4 \times 3}{51 \times 50} = \dfrac{12}{2550}$

Now, By definition of Bayes Theorem, we have

$\sf P(E_1 \mid \: E) = \dfrac{P(E_1) . P(E \mid E_1)}{P(E_1) . P(E \mid E_1) + P(E_2) . P(E \mid E_2)}$

So, on substituting the values, we get

$\sf P(E_1 \mid \: E) = \dfrac{\dfrac{1}{13} \times \dfrac{6}{2550} }{\dfrac{1}{13} \times \dfrac{6}{2550} + \dfrac{12}{13} \times \dfrac{12}{2550} }$

$\sf P(E_1 \mid \: E) = \dfrac{6}{6 + 12 \times 12}$

$\sf P(E_1 \mid \: E) = \dfrac{6}{6 + 144}$

$\sf P(E_1 \mid \: E) = \dfrac{6}{150}$

$\sf\implies \sf P(E_1 \mid \: E) = \dfrac{1}{25}$

Hence,

$\implies\sf \: \boxed{\sf \: Required\:probability\:=\:\dfrac{1}{25} \: }$