Question

a standard deck of 52 playing cards contains 13 cards in each of four suits: diamonds, hearts, clubs, and spades. two cards are chosen from the deck at random. what is the approximate probability of choosing one club and one heart? 0.0588 0.0637 0.1176 0.1275

Explanation:

Step1: Calculate the number of ways to choose one club and one heart.

There are 13 clubs and 13 hearts. The number of ways to choose 1 club from 13 is \( C(13,1)=\frac{13!}{1!(13 - 1)!}=13\), and the number of ways to choose 1 heart from 13 is also \( C(13,1) = 13\). Since the order of choosing club and heart can be club then heart or heart then club, we have two cases. But we can also think of it as the number of ways to choose one club and one heart is \(13\times13\times2\)? Wait, no. Wait, actually, when we choose two cards, one club and one heart, the number of ways is \(C(13,1)\times C(13,1)\) (because we choose 1 club from 13 and 1 heart from 13). Wait, no, the total number of ways to choose 2 cards from 52 is \(C(52,2)=\frac{52!}{2!(52 - 2)!}=\frac{52\times51}{2\times1}=1326\).

The number of favorable outcomes: number of ways to choose 1 club and 1 heart. The number of clubs is 13, number of hearts is 13. So the number of ways to choose one club and one heart is \(13\times13\) (choose a club and then a heart) plus \(13\times13\) (choose a heart and then a club)? No, actually, when we choose two cards, the number of ways to have one club and one heart is \(13\times13\times2\)? Wait, no. Wait, the number of ways to choose 1 club and 1 heart is \(C(13,1)\times C(13,1)\) because for each club, we can pair it with each heart. Wait, \(C(13,1)\) is 13, so \(13\times13 = 169\)? Wait, no, that's not right. Wait, the number of ways to choose one club and one heart is \(13\times13\) (since there are 13 clubs and 13 hearts, and each club can be combined with each heart). But the total number of ways to choose two cards is \(C(52,2)=\frac{52\times51}{2}=1326\).

Wait, let's re - calculate. The probability of choosing a club first and then a heart: the probability of choosing a club first is \(\frac{13}{52}\), then the probability of choosing a heart from the remaining 51 cards is \(\frac{13}{51}\). The probability of choosing a heart first and then a club: the probability of choosing a heart first is \(\frac{13}{52}\), then the probability of choosing a club from the remaining 51 cards is \(\frac{13}{51}\). So the total probability is \(2\times\frac{13}{52}\times\frac{13}{51}\).

Step2: Simplify the expression.

First, \(\frac{13}{52}=\frac{1}{4}\). So the expression becomes \(2\times\frac{1}{4}\times\frac{13}{51}=\frac{1}{2}\times\frac{13}{51}=\frac{13}{102}\approx0.1275\)? Wait, no, that can't be. Wait, wait, no, let's recalculate:

\(2\times\frac{13}{52}\times\frac{13}{51}=2\times\frac{1}{4}\times\frac{13}{51}=\frac{13}{102}\approx0.1275\)? Wait, but let's check the total number of favorable outcomes. The number of ways to choose 1 club and 1 heart is \(13\times13\times2\)? No, wait, the number of ways to choose two cards, one club and one heart is \(C(13,1)\times C(13,1)=13\times13 = 169\)? Wait, no, \(C(13,1)\) is 13, so 13 clubs and 13 hearts, the number of pairs is \(13\times13 = 169\)? Wait, no, when we choose two cards, the number of ways to have one club and one heart is \(13\times13\) (because for each of the 13 clubs, we can pair it with each of the 13 hearts). But the total number of ways to choose two cards is \(C(52,2)=\frac{52\times51}{2}=1326\). Then the probability is \(\frac{13\times13\times2}{52\times51}\)? Wait, no, why times 2? Wait, no, the number of ways to choose one club and one heart is \(13\times13\) (club then heart) plus \(13\times13\) (heart then club)? No, actually, when we count combinations (not permutations), the number of ways to choose one club and one heart is \(C(13,1)\times C(13,1)=13\times13 = 169\). Wait, no, combinations: the number of ways to choose 2 cards where one is club and one is heart is equal to the number of ways to choose 1 club from 13 and 1 club from 13, so \(C(13,1)\times C(13,1)=13\times13 = 169\). Then the probability is \(\frac{169\times2}{52\times51}\)? No, I'm getting confused. Let's use the formula for probability of two - step events.

The probability of first card being a club and second a heart: \(\frac{13}{52}\times\frac{13}{51}=\frac{13\times13}{52\times51}=\frac{169}{2652}\)

The probability of first card being a heart and second a club: \(\frac{13}{52}\times\frac{13}{51}=\frac{169}{2652}\)

Total probability: \(\frac{169 + 169}{2652}=\frac{338}{2652}=\frac{13}{102}\approx0.1275\)? Wait, no, \(\frac{338}{2652}=\frac{338\div26}{2652\div26}=\frac{13}{102}\approx0.1275\). Wait, but let's check with combinations. The number of ways to choose 2 cards from 52 is \(C(52,2)=\frac{52!}{2!50!}=\frac{52\times51}{2}=1326\). The number of ways to choose 1 club and 1 heart is \(13\times13 = 169\) (since we choose 1 club from 13 and 1 heart from 13). Wait, no, that's not right. Wait, the number of ways to choose 1 club and 1 heart is \(C(13,1)\times C(13,1)=13\times13 = 169\). Then the probability is \(\frac{169}{1326}\times2\)? Wait, no, because in combinations, the order doesn't matter. So the number of favorable combinations is \(13\times13 = 169\)? Wait, no, if we think of combinations, the number of ways to have one club and one heart is the number of pairs where one is club and one is heart. So for each club (13 clubs), we can pair it with each heart (13 hearts), so 13×13 = 169 combinations. Then the probability is \(\frac{169}{1326}\times2\)? No, that's wrong. Wait, no, in combinations, when we choose two cards, the number of ways to have one club and one heart is \(C(13,1)\times C(13,1)=13\times13 = 169\). Then the probability is \(\frac{169}{1326}\approx0.1275\)? Wait, \(\frac{169}{1326}=\frac{13}{102}\approx0.1275\). Yes, because 169×2 = 338, 338÷2652 = 13÷102≈0.1275.

Wait, but let's check the answer options. 0.1275 is one of the options.

Answer:

0.1275