Probability

Let E1 mean the bulb was made by unit A.
Let E2 mean the bulb was made by unit B.
Let E3 mean the bulb was made by unit C.
Let E mean the bulb is defective.
The chance a bulb comes from each unit is: P(E1)=‌

25

100

( from A),P(E2)=‌

35

100

( from B), and P(E3)=‌

40

100

( from C).
The chance a bulb is defective from each unit is:
P(‌

E

E1

)=‌

5

100

,P(‌

E

E2

)=‌

4

100

,P(‌

E

E3

)=‌

2

100

.
To find how likely it is a defective bulb came from unit B, use Bayes' theorem:
P(‌

E2

E

)=‌

P(E2)P( E E2 )

P(E1)P(‌ E E1 )+P(E2)P(‌ E E2 )+P(E3)P(‌ E E3 )

Now plug in the numbers:
P(‌

E2

E

)=‌

‌ 35 100 × 4 100

25 100 ×‌ 5 100 +‌ 35 100 ×‌ 4 100 +‌ 40 100 ×‌ 2 100

The top ( 35×4=140 ) and the bottom adds up ( 25×5=125,35×4=140, 40×2=80 ) so:
‌P(‌

E2

E

)=‌

140

125+140+80

‌P(‌

E2

E

)=‌

140

345

=‌

28

69