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Exercise 9.5 · Q5

Q.Two cards from a pack of 52 cards are lost. From the remaining cards of the pack a card is drawn at random and is found to be spade. Find the probability that the lost cards are both spades.

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Bayes' theorem over the three possibilities for how many of the 2 lost cards were spades gives P(both lost spades∣drawn is spade)=22425≈0.052P(\text{both lost spades}\mid\text{drawn is spade})=\dfrac{22}{425}\approx0.052.

Bayes' theorem with hypotheses H2,H1,H0=H_2,H_1,H_0= "22, 11, 00 of the lost cards are spades", and event E=E= "a card drawn at random from the remaining 5050 is a spade":

P(H2∣E)=P(H2) P(E∣H2)P(H2)P(E∣H2)+P(H1)P(E∣H1)+P(H0)P(E∣H0)P(H_2\mid E)=\frac{P(H_2)\,P(E\mid H_2)}{P(H_2)P(E\mid H_2)+P(H_1)P(E\mid H_1)+P(H_0)P(E\mid H_0)}

  1. Prior probabilities of how many lost cards are spades (out of 5252 cards, 1313 spades, choosing 22 lost cards):

P(H2)=(132)(522)=781326=117P(H_2)=\frac{\binom{13}{2}}{\binom{52}{2}}=\frac{78}{1326}=\frac{1}{17}

P(H1)=(131)(391)(522)=13×391326=5071326=1334P(H_1)=\frac{\binom{13}{1}\binom{39}{1}}{\binom{52}{2}}=\frac{13\times39}{1326}=\frac{507}{1326}=\frac{13}{34}

P(H0)=(392)(522)=7411326=1934P(H_0)=\frac{\binom{39}{2}}{\binom{52}{2}}=\frac{741}{1326}=\frac{19}{34}

Check: 117+1334+1934=234+1334+1934=3434=1\frac{1}{17}+\frac{13}{34}+\frac{19}{34}=\frac{2}{34}+\frac{13}{34}+\frac{19}{34}=\frac{34}{34}=1 ✓.

  1. Probability the randomly drawn card (from the remaining 5050) is a spade, under each hypothesis:

P(E∣H2)=13−250=1150(11 spades left of 50 cards)P(E\mid H_2)=\frac{13-2}{50}=\frac{11}{50}\quad(\text{11 spades left of 50 cards})

P(E∣H1)=13−150=1250P(E∣H0)=13−050=1350P(E\mid H_1)=\frac{13-1}{50}=\frac{12}{50}\qquad P(E\mid H_0)=\frac{13-0}{50}=\frac{13}{50}

  1. Compute each joint term (common denominator 1700=50×341700=50\times34):

P(H2)P(E∣H2)=117×1150=11850=221700P(H_2)P(E\mid H_2)=\frac{1}{17}\times\frac{11}{50}=\frac{11}{850}=\frac{22}{1700}

P(H1)P(E∣H1)=1334×1250=1561700P(H_1)P(E\mid H_1)=\frac{13}{34}\times\frac{12}{50}=\frac{156}{1700} …

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