Chapter 11, Problem 16E

(a)

To determine

To Explain: the simulation for a component.

Explanation of Solution

Given:

  $n\left(E\right)=C\left(13,1\right)C\left(4,2\right)$

  $n\left(S\right)=C\left(52,2\right)$

The probability that a pair is drawn $=\frac{n\left(E\right)}{n\left(S\right)}$

  $\begin{array}{l}=\frac{C\left(13,1\right)C\left(4,2\right)}{C\left(52,2\right)}\\ =0.05882\end{array}$

Required probability is

  $\begin{array}{l}=\frac{C\left(12,1\right)C\left(4,2\right)}{C\left(48,2\right)}\\ =0.06383\end{array}$

So now, for a 5-card hand, the probability of drawing two pairs and one different kind of card is given by:

The chance of drawing two pairs of a kind and in a hand with 5 cards $\begin{array}{l}=\frac{C\left(12,1\right)C\left(4,2\right)}{C\left(48,2\right)}\times \frac{C\left(12,1\right)C\left(4,2\right)}{C\left(48,2\right)}\times \frac{C\left(11,1\right)C\left(4,1\right)}{C\left(44,1\right)}\\ =0.003755\\ =0.0038\\ =\frac{38}{10000}\end{array}$

The chance of drawing three of a kind and two of some other kind in a five-card hand $\begin{array}{l}=\frac{C\left(13,1\right)C\left(4,3\right)}{C\left(52,3\right)}\times \frac{C\left(48,2\right)}{C\left(48,2\right)}\\ =0.0024\\ =\frac{24}{10000}\end{array}$

Four random digits should be used, 00-37 describing two pairs of a kind and 38-61 describing the "three of a kind" and the remaining digit (62-9999) describing the other findings (which in the current study would be of no interest).

(b)

To determine

To Explain: the simulation a trial.

Explanation of Solution

Five card hands are a trail. 10000 two-digit random numbers should be tested, and how many times each result occurs must be counted. Whatever occurs most of the time can be represented as a more probable outcome.

(c)

To determine

To Explain: the response variable.

Explanation of Solution

The response variable is if a "two pairs of a kind" of "three of a kind" is the result of a 5-card hand.