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ss7ja [257]
3 years ago
8

You may remember getting toys for Christmas or your birthdays that your parents had to assemble. One such toy requires the use o

f 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?
Physics
1 answer:
Flauer [41]3 years ago
5 0

Answer:

0.2901

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

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