69,359 views
0 votes
0 votes
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

User Kodiologist
by
2.8k points

1 Answer

16 votes
16 votes

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.

User Packie
by
2.8k points