3.11 Review Questions
 Roll five balanced dice,
 find the probability of rolling all 1s. Let A be the event that the outcomes are all 1s.
 find the probability that all the dice come up the same number.
 Are the two events in parts (a) and (b) independent?
 Are the two events in parts (a) and (b) mutually exclusive?
 If events A and B are mutually exclusive, P(A) = 0.25, P(B) = 0.4. Find the probability of each of the following events: not A, (A &B), and (A or B).
 A survey on 1000 employees about their gender and marital status gives the following data: 813 employees are male, 875 are married, and 572 married men. Is there anything wrong with these data? Explain why.
 Suppose that STAT 151 has 8 sections, 3 students each randomly pick a section. Find the probability that
 they end up in the same section.
 they are all in different sections.
 nobody picks section 1.
 There are 10 students in our class. Assume that every student is equally likely to be born on any of the 365 days in a year. Find the probability that no two students in the class have the same birthday.
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[latex]\begin{aligned} P(\mbox{all 1s})&=P(1 \mbox{ for 1st die } \& 1 \mbox{ for 2st die } \&1 \mbox{ for 3rd die } \&1 \mbox{ for 4th die } \&1 \mbox{ for 5th die})\\ &=P(1 \mbox{ for 1st die})\times P(1 \mbox{ for 2st die}) \times P(1 \mbox{ for 3rd die})\times P(1 \mbox{ for 4th die})\\&\times P(\&1 \mbox{ for 5th die}) \\ &=\frac{1}{6}\times \frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}=\left(\frac{1}{6}\right)^5=0.000128.\end{aligned}[/latex]
[latex]\begin{aligned} P(\mbox{same number})&=P(\mbox{all 1s or all 2s or all 3s or all 4s or all 5s or all 6s})\\ &=P(\mbox{all 1s})+P(\mbox{all 2s})+P(\mbox{all 3s})+P(\mbox{all 4s})+P(\mbox{all 5s})+P(\mbox{all 6s})\\&=6\times \left(\frac{1}{6}\right)^5=\left(\frac{1}{6}\right)^4=0.0007716.\end{aligned}[/latex] Events A and B are not independent, since P(A & B) = P(A) ≠ P(A) × P(B).
 Events A and B are not mutually exclusive, since the overlap is event A which is NOT an empty set.

[latex]\begin{aligned} P(\mbox{not A})&=1P(A)=10.25=0.75\\ P(A \mbox{ & } B)&=0 \quad (\mbox{mutually exclusive})\\ P(A \mbox{ or } B)&=P(A)+P(B)=0.25+0.4=0.65.\end{aligned}[/latex] There are 1000813=187 females. However, the number of married females is 875572=303 which is larger than 187 and this is impossible.

 Apply the basic counting rule:
[latex]\frac{8\times 1\times 1}{8\times 8\times 8}=\frac{1}{64}=0.015625.[/latex]  Apply the basic counting rule:
[latex]\frac{8\times 7\times 6}{8\times 8\times 8}=\frac{42}{64}=0.65625.[/latex]  Apply the basic counting rule:
[latex]\frac{7\times 7\times 7}{8\times 8\times 8}=\left(\frac{7}{8}\right)^3=0.6699.[/latex]
 Apply the basic counting rule:

[latex]\frac{ \left( \begin{array}{@{ }c@{ }} 365\\10 \end{array} \right) \times 10!}{365^{10}}=\frac{365\times 364\times \cdots\times 356}{365^{10}}=0.88305.[/latex]