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 probability of 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,

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