Question

The factory quality control department discovers
that the conditional probability of making a
manufacturing mistake in its precision ball bearing
production is 4% on Tuesday, 4% on Wednesday,
4% on Thursday, 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?​

Answer 1

The  probability that a defective ball bearing was  manufactured on a Friday is 0.375.

Step-by-step explanation:

We are given that the factory quality control department discovers  that the conditional probability of making a  manufacturing mistake in its precision ball bearing  production is 4% on Tuesday, 4% on Wednesday,  4% on Thursday, 8% on Monday, and 12% on  Friday.

The Company manufactures an equal amount of ball bearings (20%) each weekday.

Let D = event that the ball bearing is defective

So, the probability of manufacturing mistake in ball bearing  production on Monday = P(D/M) = 0.08

The probability of manufacturing mistake in ball bearing  production on Tuesday = P(D/T) = 0.04

The probability of manufacturing mistake in ball bearing  production on Wednesday = P(D/W) = 0.04

The probability of manufacturing mistake in ball bearing  production on Thursday = P(D/TH) = 0.04

The probability of manufacturing mistake in ball bearing  production on Friday = P(D/F) = 0.12

As we know that $P(A/B) = \frac{P(A\bigcap B)}{P(B)}$ , so;

$P(A\bigcap B)}={ P(A/B) \times P(B)}$

Also, the probability of ball bearing manufacturing on each weekday is 20%, that means;

P(M) = P(T) = P(W) = P(TH) = P(F) = 0.20

Now, $P(D\bigcap M)}={ P(D/M) \times P(M)}$

$P(D\bigcap M)}={ 0.08 \times 0.20$ = 0.016

$P(D\bigcap T)}={ P(D/T) \times P(T)}$

$P(D\bigcap T)}={ 0.04 \times 0.20$ = 0.008

$P(D\bigcap W)}={ P(D/W) \times P(W)}$

$P(D\bigcap W)}={ 0.04 \times 0.20$ = 0.008

$P(D\bigcap TH)}={ P(D/TH) \times P(TH)}$

$P(D\bigcap TH)}={ 0.04 \times 0.20$ = 0.008

$P(D\bigcap F)}={ P(D/F) \times P(F)}$

$P(D\bigcap F)}={ 0.12 \times 0.20$ = 0.024

So, P(D) = $P(D\bigcap M)}+P(D\bigcap T)}+P(D\bigcap W)}+P(D\bigcap TH)}+P(D\bigcap F)}$

P(D) = 0.016 + 0.008 + 0.008 + 0.008 + 0.024 = 0.064

Now, the  probability that a defective ball bearing was  manufactured on a Friday is given by = P(F/D)

            P(F/D)  =  $\frac{P(D\bigcap F)}{P(D)}$

                        =  $\frac{0.024}{0.064}$  = 0.375.