Answer:
The probability that a defective ball bearing was manufactured on a Friday = 0.375
Step-by-step explanation:
Let the event of making a mistake = M
The event of making a precision ball bearing production on Monday = Mo
The event of making a precision ball bearing production on Tuesday = T
The event of making a precision ball bearing production on Wednesday = W
The event of making a precision ball bearing production on Thursday = Th
The event of making a precision ball bearing production on Friday = F
the conditional probability of making a manufacturing mistake in its precision ball bearing production is 4% on Tuesday, P(M|T) = 4% = 0.04
4% on Wednesday, P(M|W) = 0.04
4% on Thursday, P(M|Th) = 0.04
8% on Monday, P(M|Mo) = 0.08
and 12% on Friday = P(M|F) = 0.12
The Company manufactures an equal amount of ball bearings (20 %) on each weekday, Hence, the probability that a random precision ball bearing was made on a particular day of the week, is mostly the same for all the five working days.
P(Mo) = 0.20
P(T) = 0.20
P(W) = 0.20
P(Th) = 0.20
P(F) 0.20
The probability that a defective ball bearing was manufactured on a Friday = P(F|M)
P(F|M) = P(F n M) ÷ P(M)
P(F n M) = P(M n F)
P(M) = P(Mo n M) + P(T n M) + P(W n M) + P(Th n M) + P(F n M)
We can obtain each of these probabilities by using the expression for conditional probability.
P(Mo n M) = P(M|Mo) × P(Mo) = 0.08 × 0.20 = 0.016
P(T n M) = P(M|T) × P(T) = 0.04 × 0.20 = 0.008
P(W n M) = P(M|W) × P(W) = 0.04 × 0.20 = 0.008
P(Th n M) = P(M|Th) × P(Th) = 0.04 × 0.20 = 0.008
P(F n M) = P(M|F) × P(F) = 0.12 × 0.20 = 0.024
P(M) = P(Mo n M) + P(T n M) + P(W n M) + P(Th n M) + P(F n M)
P(M) = 0.016 + 0.008 + 0.008 + 0 008 + 0.024 = 0.064
P(F|M) = P(F n M) ÷ P(M)
P(F n M) = P(M n F) = 0.024
P(M) = 0.064
P(F|M) = P(F n M) ÷ P(M) = (0.024/0.064) = 0.375
Hope this Helps!
