Chapter 13.6, Problem 35E

To determine

To find: the probability of the team to win at least $12$ out $16$ games.

Answer to Problem 35E

The probability to win at least $12$ games out of $16$ games is $44.99\%$ .

Explanation of Solution

Given:

The probability of winning team to win is $70\%$ .

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

The probability of winning team to win is $70\%$ .

The probability to winning team to lose the game is $30\%$ .

The probability to win at least $12$ games out of $16$ games is:

  $\begin{array}{l}C{}_{12}{\left(0.7\right)}^{12}{\left(0.3\right)}^4+C{}_{13}{\left(0.7\right)}^{13}{\left(0.3\right)}^3+C{}_{14}{\left(0.7\right)}^{14}{\left(0.3\right)}^2+C{}_{15}{\left(0.7\right)}^{15}{\left(0.3\right)}^1+C{}_{16}{\left(0.7\right)}^{16}{\left(0.3\right)}^0\\ \Rightarrow \mathrm{0.4499.}\\ \Rightarrow 44.99\%.\end{array}$

Hence, the probability to win at least $12$ games out of $16$ games is $44.99\%$ .