Question

A sample of 4 different calculators is randomly selected from a group containing 18 that are defective and 35 that have no defects. What is the probability that at least one of the calculators is defective?A. 0.180B. 0.810C. 0.179D. 0.821

Answer 1

We want the probability of "at least one is defective". Let's call this "D"

D -> "at least one is defective"

So we want P(D).

However, notice that the probability of "at least one is defective" plus the probability of "no on is defective" is 100%.

Let's call "no one is defective" as "N". So:

`\begin{gathered} P(N)+P(D)=1 \\ P(D)=1-P(N) \end{gathered}`

We do this because it is much easier to calculate P(N).

The probability of "N", "no one is defective", is the probability of picking one of the good ones four times in a row. We have 18 defective and 35 good ones, total of 53. So the probability of picking a good one first is:

`(35)/(53)`

Now, we have 18 defective and 34 good ones, total of 52, so the probability of picking a good one now is:

`(34)/(52)`

The third is:

`(33)/(51)`

And the fourth is:

`(32)/(50)`

The probability of all this happening is P(N), and we can calculate by:

`P(N)=(35)/(53)\cdot(34)/(52)\cdot(33)/(51)\cdot(32)/(50)\approx0.1788`

Now, we can calculate P(D):

`P(D)=1-P(N)\approx1-0.1788\approx0.821`

So, the probability os at least one of the calculator is defective is 0.821, alternative D.