Question

Judging from recent experience, 5% of the computer keyboards produced by an automatic, high-speed machine are defective. If six keyboards are randomly selected, what is the probability that none of the keyboards are defective?

Answer 1

Answer: 0.7351

Step-by-step explanation:

We use Binomial distribution here , where the probability of getting x successes in n-trials is given by :-

$P(X=x)=^nC_xp^x(1-p)^{n-x}$ , p= Probability of getting success in each trial.

As per given , the proportion of computer keyboards produced by an automatic, high-speed machine are defective : p=0.05

n=6

Let x be the number of defective keyboards.

The probability that none of the keyboards are defective will be :

$P(X=0)=^6C_0(0.05)^0(1-0.05)^{6}$

$P(X=0)=(1)(1)(0.95)^{6}\ \ [\because\ ^nC_0=1]$

$=0.735091890625\approx0.7351$

Hence, the probability that none of the keyboards are defective is probability that none of the keyboards are defective is 0.7351 .