Question

Gauri, the youngest one at home, was playing with a pack of cards. She lost a card. Later on, two cards were drawn at random from the incomplete pack. Both came out to be spade. Can you tell us the probability that lost card is a heart.

Answer 1

$\frac{13}{49?}$

Answer 2

Probability

A pack of card contains $52$ cards.

There are $4$ suits viz, Clubs, Diamonds, Hearts and Spade each having 13 denominations like $A(ace), 2,3,4,5,6,7,8,9,10, J(Jack), Q(Queen)$ and $K(King).$

There are 2 colors viz black and red — each having 26 cards.

Out of 26 red cards, 13 each from Diamonds & Hearts and out of 26 black cards, 13 each from Clubs and Spades.

E1: lost card is spade.

E2: lost card is NOT spade(heart).

A: both drawn cards are spade.

We need to find out that probability that the lost card being a heart if two cards drawn are found to be both spade.

$P(E2|A)=\frac{P(E2).P(A|E2)}{P(E1).P(A|E1)+P(E2).P(A|E2)}$

Probability that lost card is spade=$P(E1)=\frac{13}{52}=\frac{1}{4}$

P(A|E1)= Probability of getting 2 spade cards if lost card is spade=

$\frac{selecting \ spade \ cards\ from \ (13-1=12)\ cards \ from \ 51 \ cards}{selecting \ any\ two\ cards\ from \ 51 \ cards} =\frac{{12\choose 2}}{{51\choose 2}} =\frac{12\times 11}{51\times 50}$

Probability that lost card is not spade(here heart)=$1-P(E1)=1-\frac{13}{52}=1-\frac{1}{4}=\frac{3}{4}$

P(A|E2)= Probability of getting 2 spade cards if lost card is not spade(here heart)=

$\frac{selecting \ spade \ cards\ from \ 13\ cards \ from \ 51 \ cards}{selecting \ any\ two\ cards\ from \ 51 \ cards} =\frac{{13\choose 2}}{{51\choose 2}} =\frac{13\times 12}{51\times 50}$

putting the value in formula,

$P(E2|A)=\frac{P(E2).P(A|E2)}{P(E1).P(A|E1)+P(E2).P(A|E2)}$

$=\frac{\frac{3}{4}\times \frac{13\times12}{51\times50} }{\frac{1}{4}\times \frac{12\times11}{51\times50}+\frac{3}{4}\times \frac{13\times12}{51\times50}}$

$=\frac{39}{11+39}=\frac{39}{50}$

Hence, the required is  $\frac{39}{50}$.