any body can answer this quickly?

Question

two factories manufacture electric bulbs, 7% of the bulbs from factroy A and 10% of the bulbs from factory b are defective . factory B produces three times as many as bulbs as factory A each week. A bulbs is chosen at random from the week's production.

1- what is the probability that the bulb is satisfactory?

2- if the bulbs is defective, what is the probability that it came from factory A?

3- if two bulbs are chosen at random, what is the probability that they are both from factory B, knowing that the first is defective while the second is satisfactory?

Answer 1

100 fac A
300 fac B

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7 defects from fac A
30 defects from fac B

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#1

Total bulbs - 400, total defects - 37
363/400 or 90.75%

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#2

Total defects -- 37
A defects -- 7
7/37 or 18.918918...%

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#3
A defective B bulb - 30/37 or 81.081...%
A satisfactory bulb -- 270/363 or 74.ish%

The probability of both:

(30/37)x(270/363)
(8100/13431)
60.31%

Answer 2

Factory A produces 25% (1 in 4) of all the bulbs, and factory Bproduces 75% (3 in 4), so
1.
P(satisfactory) = 1 - (0.25*0.07 + 0.75*0.1) =
1 - (0.0175 + 0.075) = 1 - 0.0925 = 0.9075

2. Given a bulb is defective, probability that it came from A =
0.0175/0.0925 ≈ 0.1891892 ≈ 0.189

3. Let S = satisfactory
D = defective
P(S|B) = 0.90
P(S|A) = 0.93
P(B|S) = 0.75*0.90/0.9075 ≈ 0.7438017
P(B|D) = 0.075/0.0925 ≈ 0.8108108
P((B|S)∩(B|D) ≈ 0.6030825 ≈ 0.603

Answer 3

Say you have 400 bulbs, 100 from Factory A and 300 from Factory B. 7% from Factory A are defective, so now, you know only 393 of the bulbs can be effective. 10% of the bulbs from Factory B are defective, amounting to 30, meaning 363 of the 400 bulbs are defective. Divide that number, 363, by 4. You get 90.75% of a satisfactory bulb.

37 bulbs are defective, and 7 are from Factory A. 7/37 yields about a 19% probability it was from Factory A

30/400 results in a probability of 7.5%, and then 270/399 (because one bulb was taken) results in about 67.7%. Multiply those 2, for your answer of 5.0775% probability.