Answer
The probability of exactly one pair being dealt is 0.4226.
Work Step by Step
Consider Problem 67 in which the game is dealt with five cards from a 52-card deck. There are \[_{52}{{C}_{5}}\] ways of selecting five cards from the 52 cards. There are four cards of the same face value from which two of one face value are selected and there are 12 cards of different face values from which three cards of different face value are selected. Therefore, the number of favorable cases for one pair is \[13{{\times }_{4}}{{C}_{2}}{{\times }_{12}}{{C}_{3}}\times {{4}^{3}}\]
Now, the probability is calculated as
\[\begin{align}
& P\left( \text{exactly one pair} \right)=\frac{\text{Favorable cases}}{\text{Total cases}} \\
& =\frac{13{{\times }_{4}}{{C}_{2}}{{\times }_{12}}{{C}_{3}}\times {{4}^{3}}}{_{52}{{C}_{5}}} \\
& =\frac{13\times 6\times 220\times 64}{2,598,960} \\
& =\frac{1,098,240}{2,598,960} \\
& =0.4226
\end{align}\]
Hence, the required probability is 0.4226.
