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ynaikavde
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Factory X's bulbs work for over 5000 hours in 99% of cases 06 Mar 2015, 08:45
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C
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25% (medium)

Question Stats:

79% (02:14) correct 21% (02:38) wrong based on 115 sessions

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Factory X's bulbs work for over 5000 hours in 99% of cases, whereas factory Y's bulbs work for over 5000 hours in 95% of cases. It is known that factory X supplies 60% of the total bulbs available. What is the chance that a purchased bulb will work for longer than 5000 hours?

A) 96.4%
B) 98%
C) 97.4%
D) 95%
E) 93.9%
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C
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Factory X's bulbs work for over 5000 hours in 99% of cases 06 Mar 2015, 08:57
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ynaikavde
Factory X's bulbs work for over 5000 hours in 99% of cases, whereas factory Y's bulbs work for over 5000 hours in 95% of cases. It is known that factory X supplies 60% of the total bulbs available. What is the chance that a purchased bulb will work for longer than 5000 hours?

A) 96.4%
B) 98%
C) 97.4%
D) 95%
E) 93.9%
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hi,
probability that the purchased bulb is from factory x=3/5 and it will work=99%..
probability that the purchased bulb is from factory y=2/5 and it will work=95%..
so the prob that the bulb will work>5000h is 3/5*(99/100)+ 2/5*(95/100)..
now is the time we can save on time by avoiding tedious calculations...
difference between the two %=4..
since the ratio of number of bulbs of x:y=60:40=3:2...
we divide this gap of 4 in 5 parts and our answer will be either 3times the difference above 95 or 2 times the difference below 99..
lets take 3 parts=3*4/5=2.4... ans =95+2.4=97.4.. ans C..

by looking at examples you can narrow down choices...
answer has to be between 95 and 99 and closer to 99 as more bulbs have 99 as probability.. so only 97.4 and 98 are left..
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Factory X's bulbs work for over 5000 hours in 99% of cases 06 Mar 2015, 09:25
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Total probability
P(A) =P)(A|B1).P(B1) +P(A|B2).P(B2)

A= bulb works
B= factory from which it was bought


P(A) =P)(A|B1).P(B1) +P(A|B2).P(B2)=\(\frac{99}{100}*\frac{60}{100}+\frac{95}{100}*\frac{40}{100}=\frac{(94 + 380)}{1000}=\frac{974}{1000}\)
where

P(B1)=60/100 is the probability that the purchased bulb was manufactured by factory X;
P(B2)=4/100 is the probability that the purchased bulb was manufactured by factory Y;
P(A|B1)=99/100 is the probability that a bulb manufactured by X will work for over 5000 hours;
P(A|B2)=95/100 is the probability that a bulb manufactured by Y will work for over 5000 hours.
Thus each purchased light bulb has a 97.4% chance to work for more than 5000 hours

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Factory X's bulbs work for over 5000 hours in 99% of cases 06 Mar 2015, 13:59
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Hi All,

While this question is presented as a Probability question, it's really more of a Weighted Average question. You can approach the "math" part of it in a number of different ways (ratios, fractions, percents, decimals, etc.). Here's an approach that uses decimals.

We're given a number of facts to work with:
1) Factory X's bulbs last over 5000 hours in 99% of cases.
2) Factory Y's bulbs last over 5000 hours in 95% of cases.
3) 60% of the bulbs supplied come from Factory X.

We're asked for the probability that a randomly selected bulb will last for over 5000 hours.

TESTing VALUES can help here. Since 60% of the bulbs come from Factory X, let's say there are 5 total bulbs....

Total Bulbs = 5
Bulbs from Fact. X = 3
Bulbs form Fact. Y = 2

Now we can take the probabilities of lasting more than 5000 hours for each bulb (.99 and .95, respectively) and create a larger "weighted" equation....

[3(.99) + 2(.95)]/5 =

[2.97 + 1.90]/5 =

4.87/5 =

.974 = 97.4%

Final Answer:
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C

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Factory X's bulbs work for over 5000 hours in 99% of cases 07 Mar 2015, 04:36
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Hey,

Is this also correct?

For X, 60% of 99% will work.
For Y, 40% of 95% will work. *40% is the rest of the bulb supply in the market.

So, the probability that a purchased bulb will work is:

0.60(0.99) = .594
0.40(0.95) = 0.38

The combined probability then is 59.4 + 38 = 97.4% ANS C


* However, to be fair, there is a logical leap here, as we are assuming that the remaining 40% of the bulb supply is covered by Y, which is not directly stated in the stem.
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Factory X's bulbs work for over 5000 hours in 99% of cases 07 Mar 2015, 04:52
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pacifist85
Hey,

Is this also correct?

For X, 60% of 99% will work.
For Y, 40% of 95% will work. *40% is the rest of the bulb supply in the market.

So, the probability that a purchased bulb will work is:

0.60(0.99) = .594
0.40(0.95) = 0.38

The combined probability then is 59.4 + 38 = 97.4% ANS C


* However, to be fair, there is a logical leap here, as we are assuming that the remaining 40% of the bulb supply is covered by Y, which is not directly stated in the stem.
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hi pacifist,
an observation it is not for x 60% of 99% will work but in all of them 60 % ....
For X, 60% of 99% will work.... for x, all bulbs will work for more than 5000h in 99% cases..and these bulbs are 60% of all bulbs
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Factory X's bulbs work for over 5000 hours in 99% of cases 07 Mar 2015, 04:55
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pacifist85
Hey,

Is this also correct?

For X, 60% of 99% will work.
For Y, 40% of 95% will work. *40% is the rest of the bulb supply in the market.

So, the probability that a purchased bulb will work is:

0.60(0.99) = .594
0.40(0.95) = 0.38

The combined probability then is 59.4 + 38 = 97.4% ANS C


* However, to be fair, there is a logical leap here, as we are assuming that the remaining 40% of the bulb supply is covered by Y, which is not directly stated in the stem.

hi pacifist,
an observation it is not for x 60% of 99% will work but in all of them 60 % ....
For X, 60% of 99% will work.... for x, all bulbs will work for more than 5000h in 99% cases..and these bulbs are 60% of all bulbs
Show more

Yes sure, I agree. I wasn't sure if I had got the logic correctly.
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Factory X's bulbs work for over 5000 hours in 99% of cases 18 Nov 2023, 17:27
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