Question

The manufacturer of the disk drives in one of the well-known brands of microcomputers expects 2% of the disk drives to malfunction during the microcomputers warranty period. Calculate the probability that in a sample of 100 disk drives, that not more than three will malfunction

Answer 1

Answer:

P(not more than 3 malfunction) ≈ 0.858962

Step-by-step explanation:

Let p be probability of an individual drive malfunctioning.

Then p = 0.02.  Put q = 1-p = 0.98.

Let X be the number of malfunctioning drives in the sample of 100.

Then...

• $P(X=0) = q^{100} \approx 0.132620$
• $P(X=1) = 100pq^{99} \approx 0.270652$
• $P(X=2) = \binom{100}{2}p^2q^{98}= \frac{100\times99}{2}p^2q^{98}\approx 0.273414$
• $P(X=2) = \binom{100}{3}p^3q^{97}= \frac{100\times99\times98}{3\times2}p^3q^{97}\approx 0.182276$

So...

P(not more than 3 malfunction)

= P(X=0, 1, 2 or 3)

= P(X=0) + P(X=1) + P(X=2) + P(X=3)

≈ 0.858962