Chapter 15, Problem 17E

(a)

To determine

To find: the both are good for the probability of the first two selected.

Answer to Problem 17E

0.318

Explanation of Solution

Given:

old batteries =12

bad batteries = 5

good batteries =7

Formula used:

  $\text{Probability=}\frac{\text{Favourable}\text{Cases}}{\text{Total}\text{Cases}}$

Calculation:

Find that the first 2 batteries are likely to be successful.

Seven batteries out of twelve old are fine. A good battery chances are 7/12.

There are still 11 remaining batteries in the pile    Choose another strong 11 battery, the chance again

It's 6/11 to have another nice battery.

  $\begin{array}{c}P\left(\text{two}\text{batteries}\text{are}\text{good}\right)=\frac{7}{12}\left(\frac{6}{11}\right)\\ =\frac{42}{132}\\ =0.318\end{array}$

Thus, the required probability is 0.318

(b)

To determine

To find: The probability that at least one of the first three will work.

Answer to Problem 17E

0.955

Explanation of Solution

Given:

old batteries =12

bad batteries = 5.

good batteries =7

Calculation:

There are 5 bad battery out of 12. Therefore,

  $\begin{array}{c}P\left(\text{At least one is good}\right)=1-P\left(\text{none of the 3 is good}\right)\\ =1-\left(\frac{5}{12}\right)\left(\frac{4}{11}\right)\left(\frac{3}{10}\right)\\ =1-\frac{60}{1320}\\ =0.955\end{array}$

Thus, the required probability is 0.955

(c)

To determine

To find: the probability the 1^st four will pick all the work.

Answer to Problem 17E

0.071

Explanation of Solution

Given:

old batteries =12

bad batteries = 5

good batteries =7

Formula used:

  $\text{Probability=}\frac{\text{Favourable}\text{Cases}}{\text{Total}\text{Cases}}$

Calculation:

  $\begin{array}{l}P\left(Fourbatteriesaregood\right)=\frac{7}{12}\left(\frac{6}{11}\right)\left(\frac{5}{10}\right)\left(\frac{4}{9}\right)\\ =\frac{84}{11880}\\ =0.071\end{array}$

(d)

To determine

Find: The probability of picking 5 batteries to find one that works.

Answer to Problem 17E

0.009

Explanation of Solution

Given:

old batteries =12

bad batteries = 5

good batteries =7

Formula used:

  $\text{Probability=}\frac{\text{Favourable}\text{Cases}}{\text{Total}\text{Cases}}$

Calculation:

  $\begin{array}{l}P\left(\text{1 is working out of 5 batteries}\right)=\frac{7}{12}\left(\frac{5}{11}\right)\left(\frac{4}{10}\right)\left(\frac{3}{9}\right)\left(\frac{2}{8}\right)\\ =\frac{840}{95040}\\ =0.008838\\ =0.009\end{array}$

Thus, the probability that one of battery working out of five is 0.009