Answer:
(a) The probability that at least 13 of the next 15 motherboards pass inspection is 0.6042.
(b) On average, 1.18 motherboards should be inspected until a motherboard that passes inspection is found.
Step-by-step explanation:
Let <em>X</em> = number of motherboards that pass the inspection.
The probability that a motherboards pass the inspection is P (X) = <em>p</em> = 0.85.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The probability mass function of <em>X</em> is,
![P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7Bn%5Cchoose%20x%7Dp%5E%7Bx%7D%281-p%29%5E%7Bn-x%7D%3B%5C%20x%3D0%2C1%2C2%2C3...)
(a)
Compute the probability that at least 13 of the next 15 motherboards pass inspection as follows:
P (X ≥ 13) = P (X= 13) + P (X = 14) + P (x = 15)
![={15\choose 13}0.85^{13}(1-0.85)^{15-13}+{15\choose 13}0.85^{14}(1-0.85)^{15-14}\\+{15\choose 13}0.85^{16}(1-0.85)^{15-15}\\=0.2856+0.2312+0.0874\\=0.6042](https://tex.z-dn.net/?f=%3D%7B15%5Cchoose%2013%7D0.85%5E%7B13%7D%281-0.85%29%5E%7B15-13%7D%2B%7B15%5Cchoose%2013%7D0.85%5E%7B14%7D%281-0.85%29%5E%7B15-14%7D%5C%5C%2B%7B15%5Cchoose%2013%7D0.85%5E%7B16%7D%281-0.85%29%5E%7B15-15%7D%5C%5C%3D0.2856%2B0.2312%2B0.0874%5C%5C%3D0.6042)
Thus, the probability that at least 13 of the next 15 motherboards pass inspection is 0.6042.
(b)
Computed the expected number of motherboards should be inspected until a motherboard that passes inspection is found as follows:
![Expected\ value=\frac{1}{p}=\frac{1}{0.85}= 1.1764\approx1.18](https://tex.z-dn.net/?f=Expected%5C%20value%3D%5Cfrac%7B1%7D%7Bp%7D%3D%5Cfrac%7B1%7D%7B0.85%7D%3D%201.1764%5Capprox1.18)
Thus, on average, 1.18 motherboards should be inspected until a motherboard that passes inspection is found.