Chapter 17, Problem 17E

(a)

To determine

To find: the probability for the 1^st lefty is the 5^th individual selected.

Answer to Problem 17E

0.0745

Explanation of Solution

Given:

P=0.13

Calculation:

The required probability that the 1^st lefty is the 5^th person chosen would be mention by as follows:

  $P\left(X=0\right)={0.87}^4-{0.87}^5=0.0745$

(b)

To determine

To find: the probability that there are few lefties in the 5 people.

Answer to Problem 17E

0.502

Explanation of Solution

Given:

P=0.13

Calculation:

The required probability there are few lefties among the 5 people would be mention by as follows:

  $\begin{array}{c}=1-P\left(X=0\right)\end{array}$

  $\begin{array}{c}=1-{0.87}^5\end{array}$

  $\begin{array}{c}=0.502\end{array}$

(c)

To determine

To find: the probability that the 1^st lefty is the 2^nd or 3^rd in the group.

Answer to Problem 17E

0.2115

Explanation of Solution

Given:

P=0.13

Calculation:

The required probability that 1^st lefty is the 2^nd or3^rd person would be mention by as follows:

  $\begin{array}{c}=1-P\left(X=0\right)-P\left(X=1\right)\end{array}$

  $\begin{array}{c}=1-{0.87}^5-{0.13}^1{0.87}^0\end{array}$

  $\begin{array}{c}=0.2115\end{array}$

(d)

To determine

To find: the probability that exactly three lefties in the group.

Answer to Problem 17E

0.0166

Explanation of Solution

Given:

P=0.13

Calculation:

Required Probability there are exact3 lefties in the group

  $\begin{array}{c}P\left(X=3\right){=}^5{C}_3\times {0.13}^3\times {\left(1-0.13\right)}^2\end{array}$

  $\begin{array}{c}P\left(X=3\right)=0.0166\end{array}$

(e)

To determine

To find: the probability that minimum three lefties in the group.

Answer to Problem 17E

0.0179

Explanation of Solution

Given:

P=0.13

Calculation:

Required probability for at least 3 lefties

  $\begin{array}{c}P\left(X\ge 3\right)=1-P\left(X\le 2\right)\end{array}$

  $\begin{array}{c}P\left(X\le 2\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)\end{array}$

  $\begin{array}{c}P\left(X\le 2\right){=}^5{C}_0\times {0.13}^0\times {\left(1-0.13\right)}^5{+}^5{C}_1\times 0.1\times {0.13}^2\times {\left(1-0.13\right)}^3\end{array}$

  $\begin{array}{c}P\left(X\le 2\right)=0.9821\end{array}$

Therefore, the required probability would be 1-0.9821=0.0179

(f)

To determine

To find: the probability that not greater than three lefties in the group.

Answer to Problem 17E

0.9987

Explanation of Solution

Given:

P=0.13

Calculation:

No more than 3 lefties

  $P\left(X\le 3\right)=P\left(X\le 2\right)+P\left(X=3\right)$

From the above calculation,

  $P\left(X\le 3\right)=0.9821+0.0166=0.9987$