Problem 16 - A First Course in Probability

Question

Poker dice is played by simultaneously rolling 5 dice. Show that (a) \(P\\{\text { no two alike }\\}=.0926\) (b) \(P\\{\text { one pair }\\}=.4630\) (c) \(P\\{\text { two pair }\\}=.2315\) (d) \(P\\{\text { three alike }\\}=.1543\) (e) \(P\\{\text { full house }\\}=.0386\) (f) \(P\\{\text { four alike }\\}=.0193\) (g) \(P\\{\text { five alike }\\}=.0008\)

Step-by-Step Solution

Answer

\( \frac{P(\text{no two alike})}{P(\text{one pair})} = \frac{0.0926}{0.4630} \approx 0.20 \)

Step 1: Define the Event

Let A be the event "no two alike". This means that each dice shows a different side.

Step 2: Count the Outcomes

We have 6 choices for the first die, then 5 choices for the second (since it cannot match the first), 4 choices for the third die, and so on. To find the total outcomes for event A, we calculate \(6\times 5\times 4\times 3\times 2 = 720\).

Step 3: Compute the Probability

To find the probability, we divide the number of outcomes in A by the total outcomes: \(P(A)=\frac{720}{6^5}=0.0926\). #b) One pair#

Step 4: Define the Event

Let B be the event "one pair". This means that there is exactly one pair of numbers showing among the five dice, with the other three dice having unique outcomes.

Step 5: Count the Outcomes

First, we have \(\binom{6}{1}\) ways to pick the number that forms the pair. Next, we have \(\binom{5}{3}\) ways to pick the unique numbers for the remaining three dice, and \(3!\) ways to arrange the three remaining dice. Finally, the total outcomes for event B is \(\binom{6}{1}\cdot\binom{5}{3}\cdot 3!=600\).

Step 6: Compute the Probability

To find the probability, we divide the number of outcomes in B by the total outcomes: \(P(B)=\frac{600}{6^5}=0.4630\). #c) Two pairs#

Step 7: Define the Event

Let C be the event "two pairs". This means there are two distinct pairs of numbers showing, and the last die has a unique side different from the pairs.

Step 8: Count the Outcomes

First, we have \(\binom{6}{2}\) ways to pick the two numbers that form the pairs. Next, we have \(\binom{4}{1}\) ways to choose the unique number for the remaining die. Finally, the total outcomes for event C is \(\binom{6}{2}\cdot \binom{4}{1} \cdot 5=450\).

Step 9: Compute the Probability

To find the probability, we divide the number of outcomes in C by the total outcomes: \(P(C)=\frac{450}{6^5}=0.2315\). High school teacherSplitOptionsRatio:variables=main(Types of rollsA,Types of rollsB)=(0.0926,0.4630)