Question

Let the probability that the electrocardiograph will require repair during the warranty

period is 0.2. Determine the probability that during the warranty period out of six cardiographs:

1) no more than one will require repair; 2) at least one will require repair

Answer 1

$n=6 \\ p = 0.2 \\ q = 1 -0.2 = 0.8 \\ P(X=x)=C(n,x)p^x q^{n-x}$

1) no more than one will require repair

$P(X≤1) = P(X=0) + P(X=1) \\ = \frac{6!}{0!(10-0)!} \times 0.2^0 \times 0.8^{6-0} + \frac{6!}{1!(6-1)!} \times 0.2^1 \times 0.8^{6-1} \\ = 0.262144 + 0.393216 \\ = 0.65536$

2) at least one will require repair

$P(X≥1) = 1 -P(X<1) \\ = 1 -P(X=0) \\ = 1 - [\frac{6!}{0!(10-0)!} \times 0.2^0 \times 0.8^{6-0}] \\ = 1 -0.262144 \\ = 0.737856$