Question

You may remember getting toys for Christmas or your birthdays that your parents had to assemble. One such toy requires the use of 18 bolts. Since the manufacturer knows that he produces some defective bolts (about 6 % of his production), each package contains two extra bolts, i.e., 20. If this particular package contains one defective bolt, what is the chance that your parents will not encounter a defective bolt while assembling your toy?

Answer 1

0.2901

Step-by-step explanation:

P(Defective Bolt) = 0.06

P(Non-Defective Bolt) = 0.94

n = 20

Let X be the random variable which indicates the number of defective bolts and this follows a Binomial Distribution

Therefore:

`P(X=0)= C(20,0)*(0.94)^20 * (0.06)^0 = 0.2901`