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Chapter 2: Problem 16

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) \(P\\{\text { five alike }\\}=.0008\)

Short Answer

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

Step by step solution

01

Define the Event

Let A be the event "no two alike". This means that each dice shows a different side.
02

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 \(6\times 5\times 4\times 3\times 2 = 720\).
03

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#
04

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.
05

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 \(\binom{6}{1}\cdot\binom{5}{3}\cdot 3!=600\).
06

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#
07

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.
08

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 \(\binom{6}{2}\cdot \binom{4}{1} \cdot 5=450\).
09

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)

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