Chapter 17, Problem 10E

(a)

To determine

To find: The probability of the fifth chip test being the first bad one to find.

Answer to Problem 10E

0.0184

Explanation of Solution

Given:

Model is geometric probability

Success = get good chip.

Failure = get bad chip.

  $\begin{array}{c}p=0.98\\ q=0.2\end{array}$

Formula used:

  $P\left(X=x\right)={\left(1-p\right)}^{x-1}p$

Calculation:

suppose X to be tested before bad chips have been obtained.

  $\begin{array}{c}P\left(X=5\right)=p^{4}q\\ ={\left(0.98\right)}^4\left(0.02\right)\\ =0.0184\end{array}$

(b)

To determine

To find: the probability for the first 10 tests to detect a poor one.

Answer to Problem 10E

18.3%

Explanation of Solution

Given:

Model is geometric probability

Success = get good chip.

Failure = get bad chip.

  $\begin{array}{c}p=0.98\\ q=0.2\end{array}$

Formula used:

  $P\left(X=x\right)={\left(1-p\right)}^{x-1}p$

Calculation:

X to be a reasonable amount of chips taken before the first bad chip.

It is possible that you can find a chip in the first 10,

  $\begin{array}{c}P\left(X=1\right)+\mathrm{...}+P\left(X=10\right)=\left[\begin{array}{l}\left(0.02\right)+\left(0.98\right)\left(0.02\right)+{\left(0.98\right)}^2\left(0.02\right)+\\ {\left(0.98\right)}^3\left(0.02\right)+{\left(0.98\right)}^4\left(0.02\right)+{\left(0.98\right)}^5\left(0.02\right)+\\ {\left(0.98\right)}^6\left(0.02\right)+{\left(0.98\right)}^7\left(0.02\right)+{\left(0.98\right)}^8\left(0.02\right)+\\ {\left(0.98\right)}^9\left(0.02\right)\end{array}\right]\\ =\left[\begin{array}{l}0.02+0.0196+0.0192+0.01882+\\ 0.01844+0.01808+0.01772+\\ 0.01736+0.01702+0.01668\end{array}\right]\\ =0.183\end{array}$

There is 18.3% possibility that first bad one obtains with first 10 test.