The factory quality control department discovers that the conditional probability of making a manufacturing mistake in its precision ball bearing production is 4\%4% on Tuesday, 4\%4% on Wednesday, 4\%4% on Thursday, 8\%8% on Monday, and 12\% on Friday. The Company manufactures an equal amount of ball bearings (20\ %) on each weekday. What is the probability that a defective ball bearing was manufactured on a Friday

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!

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]