JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Probability

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A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :

(A) $\frac{22}{425}$
(B) $\frac{52}{867}$
(C) $\frac{3}{4}$
(D) $\frac{39}{50}$ ✓ Correct

Answer: (D)

Solution

Let

E

= event that two drawn cards are spades,

A

= missing card is not a spade. We need

P (A ∣ E)

. By Bayes' theorem,

P (A ∣ E) = \frac{P ( E ∣ A ) P ( A )}{P ( E )}

P (A) = \frac{39}{52}

(39 non-spades out of 52). If missing card is not a spade, there are 13 spades in the remaining 51 cards.

P (E ∣ A) = \frac{( 2 13 )}{( 2 51 )} = \frac{78}{1275}

. If missing card is a spade,

P (A^{c}) = \frac{13}{52} = \frac{1}{4}

, and

P (E ∣ A^{c}) = \frac{( 2 12 )}{( 2 51 )} = \frac{66}{1275}

P (E) = P (E ∣ A) P (A) + P (E ∣ A^{c}) P (A^{c}) = \frac{78}{1275} \cdot \frac{39}{52} + \frac{66}{1275} \cdot \frac{13}{52} = \frac{78 \cdot 39 + 66 \cdot 13}{1275 \cdot 52} = \frac{3042 + 858}{66300} = \frac{3900}{66300} = \frac{1}{17}

P (A ∣ E) = \frac{\frac{78 \cdot 39}{1275 \cdot 52}}{\frac{1}{17}} = \frac{78 \cdot 39 \cdot 17}{1275 \cdot 52} = \frac{3042}{3900} = \frac{39}{50}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Probability

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