Question

A card from a pack of 52 cards is lost. From the remaining cards of pack, two cards are drawn and are found to be diamonds. Find the probability of the missing card to be diamond.

Answer 1

Answer:

$\boxed{\sf \: \: Required\:probability\:=\:\dfrac{11}{50} \: }$

Step-by-step explanation:

Let assume the following events

E₁ : Card lost is of diamond.

E₂ : Card lost is not of diamond

E : getting two diamond cards.

Now,

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

$\sf \: P(E_2) = \dfrac{39}{52} = \dfrac{3}{4}$

Now,

$\sf \: P(E \: \mid \: E_1) = \dfrac{C(12,2)}{C(51,2)} = \dfrac{12 \times 11}{51 \times 50} = \dfrac{132}{2550}$

and

$\sf \: P(E \: \mid \: E_2) = \dfrac{C(13,2)}{C(51,2)} = \dfrac{13 \times 12}{51 \times 50} = \dfrac{156}{2550}$

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

$\sf P(E_1 \mid \: E) = \dfrac{\dfrac{1}{4} \times \dfrac{132}{2550} }{\dfrac{1}{4} \times \dfrac{132}{2550} + \dfrac{3}{4} \times \dfrac{156}{2550} }$

$\sf P(E_1 \mid \: E) = \dfrac{132}{132 + 3 \times 156}$

$\sf P(E_1 \mid \: E) = \dfrac{132}{132 + 468}$

$\sf P(E_1 \mid \: E) = \dfrac{132}{600}$

$\sf\implies \sf P(E_1 \mid \: E) = \dfrac{11}{50}$

Hence,

$\implies\sf \: Required\:probability\:=\:\dfrac{11}{50}$