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Bunuel
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A watch manufacturer has two factories (FA, FB) and 60% of their watch 31 Jan 2019, 03:06
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56% (01:47) correct 44% (02:07) wrong based on 113 sessions

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A watch manufacturer has two factories (FA, FB) and 60% of their watches are made at FA. It is known that 10% of them are made at FA and 15% made at FB are defective. What is the probability that a selected defective watch was manufactured at FB?

A. 0.05
B. 0.06
C. 0.066
D. 0.11
E. 0.5
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E
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A watch manufacturer has two factories (FA, FB) and 60% of their watch Updated on: 31 Jan 2019, 09:01
Originally posted by Archit3110 on 31 Jan 2019, 03:12.
Last edited by Archit3110 on 31 Jan 2019, 09:01, edited 1 time in total.
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Bunuel
A watch manufacturer has two factories (FA, FB) and 60% of their watches are made at FA. It is known that 10% of them are made at FA and 15% made at FB are defective. What is the probability that a selected defective watch was manufactured at FB?

A. 0.05
B. 0.06
C. 0.066
D. 0.11
E. 0.5
Show more

total watches be 100
watch made at FA= 60 and FB= 40
defective watch FB= 40*.15= 6
so P defective watch FB = 6/100 = .06


Total defective watches A 6 & B 6 : 12

So question has asked a selected defective watch if selected would be of B
6/12 = 0.5 IMO E
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A watch manufacturer has two factories (FA, FB) and 60% of their watch 31 Jan 2019, 05:00
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Bunuel
A watch manufacturer has two factories (FA, FB) and 60% of their watches are made at FA. It is known that 10% of them are made at FA and 15% made at FB are defective. What is the probability that a selected defective watch was manufactured at FB?

A. 0.05
B. 0.06
C. 0.066
D. 0.11
E. 0.5
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Dear GMATGuruNY

Can you please share your thoughts on this question?

Thanks
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A watch manufacturer has two factories (FA, FB) and 60% of their watch 31 Jan 2019, 05:06
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Bunuel
A watch manufacturer has two factories (FA, FB) and 60% of their watches are made at FA. It is known that 10% of them are made at FA and 15% made at FB are defective. What is the probability that a selected defective watch was manufactured at FB?

A. 0.05
B. 0.06
C. 0.066
D. 0.11
E. 0.5
Show more

total watches be 100
watch made at FA= 60 and FB= 40
defective watch FB= 40*.15= 6
so P defective watch FB = 6/100 = .06


Total defective watches A 6 & B 6 : 12

So question has asked a selected defective watch if selected would be of B
6/12 = 0.5 IMO E

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A watch manufacturer has two factories (FA, FB) and 60% of their watch 31 Jan 2019, 05:18
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Bunuel
A watch manufacturer has two factories (FA, FB) and 60% of their watches are made at FA. It is known that 10% of them are made at FA and 15% made at FB are defective. What is the probability that a selected defective watch was manufactured at FB?

A. 0.05
B. 0.06
C. 0.066
D. 0.11
E. 0.5
Show more

Let the total number of watches = 100, implying that the number made at FA = 60% of 100 = 60 and that the number made at FB = 100-60 = 40.
Total number of defective watches = (10% of the 60 FA watches) + (15% of the 40 FB watches) = 6 + 6 = 12.
Since 1/2 of the defective watches are made at FB, the probability that a selected defective watch is made at FB = 0.5.

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E
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A watch manufacturer has two factories (FA, FB) and 60% of their watch 31 Jan 2019, 05:27
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Bunuel
A watch manufacturer has two factories (FA, FB) and 60% of their watches are made at FA. It is known that 10% of them are made at FA and 15% made at FB are defective. What is the probability that a selected defective watch was manufactured at FB?

A. 0.05
B. 0.06
C. 0.066
D. 0.11
E. 0.5

Let the total number of watches = 100, implying that the number made at FA = 60% of 100 = 60 and that the number made at FB = 100-60 = 40.
Total number of defective watches = (10% of the 60 FA watches) + (15% of the 40 FB watches) = 6 + 6 = 12.
Since 1/2 of the defective watches are made at FB, the probability that a selected defective watch is made at FB = 0.5.

Show Spoiler
E
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Thanks GMATGuruNY

But here is a question, why do not we make as follows:

Probability to pick a defective part of FA = 6/60

Probability to pick a defective part of FB = 6/40

Probability to pick a good part of FA = 54/60

Probability to pick a defective part of FB = 6/40

probability to select defective watch = (Probability to pick a defective part of FA)*(Probability to pick a defective part of FB)+(Probability to pick a good part of FA) * (Probability to pick a defective part of FB)

(6/60)(6/40) + (54/60)(6/40) = 6/40 =0.15

Where did I go wrong??

Thanks in advance
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A watch manufacturer has two factories (FA, FB) and 60% of their watch 31 Jan 2019, 08:14
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Solution


Given:
    • 60% of the watches are made at FA
    • 10% of the watches made at FA are effective, and
    • 15% of the watches made at FB are defective

To find:
    • The probability that the selected defective watch was manufactured at FB

Approach and Working:
Let the total number of watches be 100x
    • 60x are manufactured at FA,
      o 10% of 60x = 6x watches are defective

    • 40x are manufactured at FB
      o 15% of 40x = 6x watches are defective

    • Thus, total number of defective watches = 6x + 6x = 12x

Therefore, probability that selected defective watch was manufactured at FB = \(\frac{6x}{12x} = \frac{1}{2}\)

Hence, the correct answer is Option E.

Answer: E

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A watch manufacturer has two factories (FA, FB) and 60% of their watch 31 Jan 2019, 09:00
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Mo2men
[Thanks GMATGuruNY

But here is a question, why do not we make as follows:

Probability to pick a defective part of FA = 6/60

Probability to pick a defective part of FB = 6/40

Probability to pick a good part of FA = 54/60

Probability to pick a defective part of FB = 6/40

probability to select defective watch = (Probability to pick a defective part of FA)*(Probability to pick a defective part of FB)+(Probability to pick a good part of FA) * (Probability to pick a defective part of FB)

(6/60)(6/40) + (54/60)(6/40) = 6/40 =0.15

Where did I go wrong??

Thanks in advance
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P(selecting a defective part from FA) = P(selecting a part from FA) * P(the selected part from FA Is defective) = \(\frac{60}{100} * \frac{10}{100} = \frac{6}{100}\) = 0.06
P(selecting a defective part from FB) = P(selecting a part from FB) * P(the selected part from FB Is defective) = \(\frac{40}{100} * \frac{15}{100} = \frac{6}{100}\) = 0.06
Thus:
P(selecting a defective part) = 0.06 + 0.06 = 0.12

The value in blue represents ALL possible ways of selecting a defective part.
The value in green represents the GOOD ways of selecting a defective part (a defective part from FB).
Thus:
(good ways)/(all possible ways) =\(\frac{0.06}{0.12} = \frac{1}{2}\) = 0.5
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