Question

What is the probability that a random five-card hand has (a) Exactly one pair (no three of a kind or two pairs)? Comment on the answers (a) 13 × C(4,2) × C(48,3)/C(52,5), (b)13×C(4,2)×48×44×40/C(52,5), and(c)(52×3×48×44×40)/ (52×51×50×49×48).

Answer 1

Answer:

So, the probability for exactly one pair is P=0.42.

Step-by-step explanation:

We calculate the number of possible combinations to choose 5 from 52 cards.

$C_5^{52}=\frac{52!}{5!(52-5)!}=2598960$

In order to have exactly one pair, we must fulfill the following conditions.

We know that there are 13 different numbers in the card deck, with four numbers each.

So we have 13 options to choose one number, then from 4 cards we choose two.  

In the continuation of the remaining 12 numbers, we dial 1 different number to get exactly one pair. And for each of the three numbers we choose we have 4 possibilities.

So, the number of favorable combinations looks like this:

$13\cdot C^4_2\codt C_3^{12}\cdot 4^3=13\cdot \frac{4!}{2!(4-2)!}\cdot\frac{12!}{3!(12-3)!}\cdot64=1098240$

Therefore, the probability is

$P=\frac{1098240}{2598960}\\\\P=0.42$

So, the probability for exactly one pair is P=0.42.