Question

A sample of 4 different calculators is randomly selected from a group containing 17 that are defective and 37that have no defects. What is the probability that at least one of the calculators is defective?

Answer 1

Answer: 0.8025

Explanation:

Given : The number of defective calculators : 17

The number of calculators are not defective : 37

Total calculators : 37+17=54

The probability of the calculators are defective :
`(17)/(54)=(1)/(3)`

Binomial distribution formula :-

`P(x)=^nC_xp^x(1-p)^(n-x)`, where P(x) is the probability of success in x trials, n is total trials and p is the probability of success for one trial.

The probability that at least one of the calculators is defective is given by :-

`P(x\geq1)=1-P(0)\\\=1-(^4C_0((1)/(3))^0(1-(1)/(3))^4)\\\\=1-((2)/(3))^4=0.80246913\approx0.8025`