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What is the probability of being dealt 3 tens and 2 kings from a deck of cards?

A. [tex]$\frac{3}{245,162}$[/tex]
B. [tex]$\frac{1}{108,290}$[/tex]
C. [tex]$\frac{5}{52}$[/tex]
D. [tex]$\frac{1}{52}$[/tex]


Sagot :

To determine the probability of being dealt 3 tens and 2 kings from a standard deck of 52 cards, let's go through the steps carefully:

1. Total Number of Ways to Choose 5 Cards Out of 52:
We need to calculate how many ways we can choose any 5 cards from a deck of 52 cards. This is a combination problem, and the formula for combinations is given by:
[tex]\[ \binom{52}{5} = \frac{52!}{5!(52-5)!} \][/tex]

2. Number of Ways to Choose 3 Tens Out of 4:
There are 4 tens in a deck (ten of hearts, ten of diamonds, ten of clubs, ten of spades). We need to find out the number of ways to choose 3 out of these 4 tens:
[tex]\[ \binom{4}{3} = \frac{4!}{3!(4-3)!} = 4 \][/tex]

3. Number of Ways to Choose 2 Kings Out of 4:
Similarly, there are 4 kings in a deck. We need to find the number of ways to choose 2 out of these 4 kings:
[tex]\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = 6 \][/tex]

4. Number of Favorable Outcomes:
The number of favorable ways to choose 3 tens and 2 kings simultaneously is the product of the combinations of tens and kings:
[tex]\[ \text{Favorable Ways} = \binom{4}{3} \cdot \binom{4}{2} = 4 \cdot 6 = 24 \][/tex]

5. Probability:
Finally, the probability of being dealt 3 tens and 2 kings is the ratio of favorable outcomes to the total number of possible outcomes:
[tex]\[ \text{Probability} = \frac{\text{Favorable Ways}}{\binom{52}{5}} \][/tex]

Given the answer from calculations, the probability is approximately:
[tex]\[ 9.23446301597562 \times 10^{-6} \][/tex]

6. Matching Probability to the Given Choices:
The numerical value [tex]\(9.23446301597562 \times 10^{-6}\)[/tex] is equivalent to [tex]\(\frac{1}{108,290}\)[/tex] when expressed as a simplified fraction.

Thus, the best answer for the question is:
[tex]\[ \boxed{\frac{1}{108,290}} \][/tex]

So, the correct option is:
B. [tex]\(\frac{1}{108,290}\)[/tex]