A box of light bulbs contains exactly 3 light bulbs that are defective

from the given, we know the number of defective bulbs,
but we don't know:
a- the number of non-defective bulbs
b- the number of the sample to be picked.
c- the method of picking (with or without replacement)

statement (1) provides us with info (c) only --> insufficient
statement (2) provides us with info (b) only --> insufficient

so both (1) and (2) are insufficient because they can't provide info (a)

for example, if the number of non-defective bulbs = 17, so the probability will be 1
and as the number of non-defective bulbs increase, the probability starts to decrease below 1

Straight forward: Since we do not know the sample size (total number of bulbs to be picked), we cannot know the probability of getting atleast 1 defective bulb

How could not we know the info a? As the stimulus provides us with "exactly 3 defective bulbs" so i assume the rest are non-defective bulbs. What am i wrong here? Please shed some lights. Thank you

We can't find the total number of bulbs in the box from statement 1 or 2. Hence, it's impossible to tell the probability that a sample of light bulbs picked will contain atleast 1 defective light bulb

If total number of bulbs are 21, probability that probability that a sample of 20 light bulbs picked will contain atleast 1 defective light bulb is equal to 1.
If total number of bulbs are 23, probability that probability that a sample of 20 light bulbs picked will contain atleast 1 defective light bulb is less than 1.

Question stem- No.of defective light bulbs are 3.
Total no.of bulbs= unknown

We are asked to find -the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb.

We need to know total no.of bulbs present in the box and picking methods.

statement 1- Insufficient
Total no.of bulbs not known

Statement 2-insufficient
Picking method not known. We don't know whether 1 bulb is picked at a time, 2 picked at a time or with/without replacement.

.
Statement 1 + Statement 2

Insufficient.

Reason is we cannot arrive at a unique answer because question stem asks to find probability that atleast 1 defective bulb is picked.
There are 3 cases possible here-
one is defective, 2 defective, 3 are defective. The probability of these 3 cases gives different answer. Hence correct option is E.
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Given :-
i) There is a box of bulbs that has 3 defective bulbs. But we do not know the total number of bulbs, say n.
ii) We pick up certain bulbs, and are looking for the probability of at least 1 in them to be defective. Again we do not know the size of sample, say x.

Statements :-

(1) The light bulbs in the sample will be picked 1 at a time without replacement.
Ok, but we are looking for the total bulbs and number of bulbs picked up as sample.
Insuff

(2) The sample will contain exactly 20 light bulbs.
x=20, but what about n. Insuff

Combined.
If n=x=20... Probability =1
But if n= 40, and x=20, probability <1.
Insuff

E
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You don't know what you are talking about bro. The reason why 1+2 is insfficient is that we don't know how many bulbs we pick. If we pick only one, the probability will be 3/20; If we pick two, the probability will be 27/95...

The reason why (2) is insufficient is also that we don't know how many bulbs we pick. it doesn't matter if we put the bulbs back or not because we are picking bulbs, not picking letters. In other words, we are looking at combination not arrangement!!

Let's have a go at it in a structured manner.
Given:
Box has some number of bulbs(let that number be x)
Out of x bulbs, 3 are defective.

What we have to find:
Probability that out of a sample(let size of sample be y) of bulbs, at least 1 bulb is defective.

Process above info:
So we are given x bulbs, out of which 3 bulbs are defective and we need to choose a sample of y bulbs.
We have to find the probability that at least one of the bulbs in sample of size y is defective.
P(at least 1 defective) = 1 - P(no bulb defective)

Now moving on to the statements:
1)It tells that we pick bulbs one by one without replacement. This is similar to picking all y bulbs at once. But statement 1) does not provide any info about total bulbs(x) or sample size(y). So not sufficient to find probability.
Not Sufficient

2) This tells us that y=20. But still no info about x.
Not sufficient

1) and 2) combined:
Combined they tell us that we pick bulbs one by one w/o replacement and y=20. Still no information about x.
Hence Not Sufficient
Answer is E

Statement 2 states The sample will contain exactly 20 light bulbs, it does not give us the total number of bulbs.
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To Find: Probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb

Know: A box of light bulbs contains exactly 3 light bulbs that are defective.

Missing information:
1) no. of non-defective bulbs
2) the number of bulbs in the sample
3) the method of picking (with or without replacement)

St (1) tells us about the method of picking (without replacement). We don't know about the other 2 missing pieces.
INSUFFICIENT

St (2) tells us about the number of bulbs in the sample. We don't know about the other 2 missing pieces.
INSUFFICIENT

Combining the information from 2 statements give us 2 missing information but do not tell us about the number of defective bulbs.
INSUFFICIENT

Answer - E

1+2

So if the box had 500 box and we're picking a sample of 20 bulbs from the 500 and checking the P(of atleast one defective)

1c20, 2c20, 3c20/ 20c500

\(1c20, 2c20, 3c20\) :- 1, 2 ,3 out of the 20 picked are defective.

but in the denominator, we have no idea what's the total number of bulbs, here we have taken 500 so we will get some results.

but if we took n= 1000, that denominator will become, \(20C1000 \). A world of difference.

hence, E.

Correct option : E

We need concreate 4 information to solve
1. Total Bulb quantity sample - variable 1 (Available)
2. Distinct things occuring at set interval times - variable 2 (Not available)
3. Selection range sample - variable - 3 (Not available)
4. Events defective no's - variable 4 (Available)

based on below information:
A box of light bulbs contains exactly 3 light bulbs that are defective. What is the probability that a sample of light bulbs picked at random from this box will contain at least 1 defective light bulb?
(1) The light bulbs in the sample will be picked 1 at a time without replacement.
(2) The sample will contain exactly 20 light bulbs

A, B, D rejected
C rejected

Winner E

Dear Expert,

From the probability formula P(E) = n(E)/n(S)

the first statement, we know the picking method.
The second statement, we only know that we picked 20 bulbs from the total bulbs that we don't know how many bulbs we have.

Combine (1) and (2), we don't know n(s) - which is total bulbs. We do know that we will pick 1 bulb 20 times without replacement
So, we cannot calculate the P(E)

Am I correct?

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