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Chapter 2: Problem 112

A hand of five cards is to be dealt at random and without replacement from an ordinary deck of 52 playing cards. Find the conditional probability of an all spade hand given that there will be at least 4 spades in the hand.

Short Answer

Expert verified
The conditional probability of getting an all-spade hand given at least 4 spades in the hand is approximately \(\frac{1287}{8443}\), or approximately \(0.1524\), or 15.24%.

Step by step solution

01

Calculate the total number of ways to get 4 spades and 1 non-spade

To calculate the total number of ways to get 4 spades and 1 non-spade, we need to find the number of ways to choose 4 spades out of the 13 in the deck, and then multiply this by the number of ways to choose 1 non-spade out of the remaining 39 cards (52 cards in the deck minus 13 spades). We can do this using combinations: \( C(13,4) \times C(39,1) \)
02

Calculate the number of ways to get all 5 spades

To calculate this, we simply need to find the number of ways to choose 5 spades out of the 13 available. We can also use combinations for this: \( C(13,5) \)
03

Calculate the total number of possible hands with at least 4 spades

Now, we need to add together the results from Step 1 and Step 2 to find the total number of possible hands with at least 4 spades: \( C(13,4) \times C(39,1) + C(13,5) \)
04

Calculate the conditional probability of an all-spade hand given at least 4 spades

Finally, we can find the conditional probability by dividing the number of ways to get all 5 spades (from Step 2) by the total number of possible hands with at least 4 spades (from Step 3): \( \frac{C(13,5)}{C(13,4) \times C(39,1) + C(13,5)} \) Now, we can plug in the values from the combination formulas: \( \frac{C(13,5)}{C(13,4) \times C(39,1) + C(13,5)} = \frac{\binom{13}{5}}{\binom{13}{4} \times \binom{39}{1} + \binom{13}{5}} \) \( = \frac{1287}{1287 + 7156} = \frac{1287}{8443} \) So the conditional probability of an all-spade hand given at least 4 spades in the hand is approximately \(\frac{1287}{8443}\), or approximately \(0.1524\), or 15.24%.

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