Question

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

Answer 1

Let E1 be the event that the lost card is a hearts. As there are 13 hearts in a deck , P(E1)=13/52
let E2 be the event that the lost card is not a hearts, P(E2)= 1-13/52 = 3/4
Let A be the event that both cards drawn are found to be of hearts.                 Now we have to calculate P( two cards are both hearts and the lost card is also of hearts).
Now 12 cards of hearts is left from the given 13 cards of hearts and total cards left are now 51.
the two cards that are both hearts can be drawn in 12C2 ways= 12*11/2*1 = 132/2 ways = 66 ways.
similarly the two cards of hearts can be drawn from the pack of 51 cards is 51C2 = 51*50/2*1 ways = 2550/2 = 1275ways.
therefore the P(the two cards drawn are hearts given one is lost P(A/E1) = 66/1275
now consider that the two cards os heart and the lost card is not hearts
the two cards that can be both hearts can be drawn in 13C2 ways = 13*12/2*1 = 156/2 ways = 78 ways
similarly the two cards can be drawn from 51 cardsin 51C2 ways = 51*50/2*1 = 2550/2 = 1275 ways.
therefore the P ( two cards drawn are hearts given that is lost is not a hearts) = p(A/E1) = 78/1225
Now by using Bayes theorem, P(E1/A) = [P(E1)*P(A/E1)]/P(E1)*P(A/E1) + P(E2)*P(A/E2) =
[66/1225 * 1/4] / {66/1225 * 1/4} + {78/1225 * 3/4} =
66/300 = 11/50.
hope my answer is correct

Answer 2

Answer:

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

Step-by-step explanation:

Let assume the following events

E₁ : Card lost is of heart.

E₂ : Card lost is not of heart

E : getting two heart 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,

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