3.38

Male children. While it is often assumed that the probabilities of having a boy or a girl are the same, the actual probability of having a boy is slightly higher at 0.51. Suppose a couple plans to have 3 kids.

  1. Use the binomial model to calculate the probability that two of them will be boys.
  2. Write out all possible orderings of 3 children, 2 of whom are boys. Use these scenarios to calculate the same probability from part (a) but using the addition rule for disjoint outcomes. Confirm that your answers from parts (a) and (b) match.
  3. If we wanted to calculate the probability that a couple who plans to have 8 kids will have 3 boys, briefly describe why the approach from part (b) would be more tedious than the approach from part (a).

  1. \[(\begin{matrix} 3 \\ 2 \end{matrix}){ 0.51 }^{ 2 }*0.49=0.382\]

  2. BBG,BGB,GBB,GGB,GBG,BGG,GGG,BBB \[P(2\quad boy)=\quad \frac { BBG+GBB+BGB }{ BBG+BGB+GBB+GGB+GBG+BGG+GGG+BBB } =\frac { 3 }{ 8 } =0.375\]

So since boy and girl has slight different probablity, two answer will be different but they are close

  1. Because using approach, you will have to list all 56 combinations

3.39

Calculate the following probabilities and indicate which probability distribution model is appropriate in each case. You roll a fair die 5 times. What is the probability of rolling

  1. the first 6 on the fifth roll?
  2. exactly three 6s?
  3. the third 6 on the fifth roll?

  1. \[P(the\quad first\quad 6\quad on\quad fifth\quad roll)=\quad { \frac { 5 }{ 6 } }^{ 4 }\frac { 1 }{ 6 } =0.803\] geometric distribution

  2. \[P(three\quad 6)=\quad { (\begin{matrix} 5 \\ 3 \end{matrix})\frac { 5 }{ 6 } }^{ 2 }{ \frac { 1 }{ 6 } }^{ 3 }=0.032\] Binomial

  3. \[P(the\quad third\quad on\quad 5th)=\quad { (\begin{matrix} 4 \\ 2 \end{matrix})\frac { 5 }{ 6 } }^{ 2 }{ \frac { 1 }{ 6 } }^{ 3 }=0.0193\] Negitve Binomial