Question

in the mass production of bolts it is found that 5% are defective. bolts are selected at random and put into packets of 10.

a. if a packet is selected at random ,find the probability that it will contain

three defective bolts

less than three defective bolts

b. two packets are selected at random . find the probability that there are no defective bolts in either packet

(SOLVE USING BINOMIAL DISTRIBUTION )

Answer 1

a)
Here, n = 10, p = 0.05, (1 - p) = 0.95 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 3)
P(X = 3) = 10C3 * 0.05^3 * 0.95^7
P(X = 3) = 0.0105

less than 3
P(X <= 2) = (10C0 * 0.05^0 * 0.95^10) + (10C1 * 0.05^1 * 0.95^9) + (10C2 * 0.05^2 * 0.95^8)
P(X <= 2) = 0.5987 + 0.3151 + 0.0746
P(X <= 2) = 0.9884

b)
We need to calculate P(X = 0)
P(X = 0) = 10C0 * 0.05^0 * 0.95^10
P(X = 0) = 0.5987

P(no defective in both) = 0.5987^2 = 0.3584