In nine independent trials, what is the probability that Outcome A

Probability of Favorable Outcome + Probability of Unfavorable Outcome = 1

i.e. Probability of Favorable Outcome = 1 - Probability of Unfavorable Outcome

i.e. probability that Outcome A happens at least once = 1 - Probability that Outcome A never happens

Statement 1: P (A never happens) = 0.026
i.e. P (A happens atleast once) = 1- 0.026 = 0.974
SUFFICIENT

Statement 2: probability of Outcome A resulting in a single trial is 1/3
i.e. probability of Outcome A Not happen in a single trial is 1-(1/3) = 2/3
i.e. Outcome A never happens in all 9 trials = (2/3)^9 = 0.026
i.e. P (A happens atleast once) = 1- 0.026 = 0.974
SUFFICIENT

Answer: Option

Probability (Outcome A happens at least once)+probability(Outcome A does not happen even once)=1.
Statement 1= Sufficient.
statement 2= We know probability of A=1/3, p(A')=2/3 with these we can find out all the cases when A occurs at least once.sufficient.

ans=> D

Here we go:

St1: The probability that Outcome A does not happen even once in any of the nine trials is 0.026

We have been asked the probability that Outcome A happens at least once.

This can be calculate ---> (1 - 0.026)

Hence St1 is sufficient

St2 : the probability of Outcome A resulting in a single trial is 1/3.
We can also calculate the probability of Outcome A not resulting in a single trial : (1 - 1/3) = 2/3

So we can find easily calculate the probability that Outcome A happens at least once by using above information.

Hence St2 is also sufficient

Option D is correct.

MAGOOSH OFFICIAL SOLUTION

We know there are nine trials. To figure out the probability that Outcome A happens at least once, we would need a way to figure out the probability of A, P(A).

Statement #1: this statement gives us the complement. The complement of (Outcome A happens at least once) is (Outcome A does not happen at all). Complementary probabilities have a sum of 1 — P(not K) = 1 – P(K). Therefore, using the complement rule, we could figure out that
P(A happens at least once) = 1 – P(A doesn’t happen at all) = 1 – 0.026 = 0.974
This information allows us to answer the prompt question. This statement, alone and by itself, is sufficient.

Statement #2: this statement gives us P(A), which would allow us to calculate the probability that A happens at least once in nine trials. This statement, alone and by itself, is sufficient.

Both statement sufficient. Answer = D

- See more at: https://magoosh.com/gmat/2013/gmat-data- ... JzRXH.dpuf
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Ans Is D coz of the following Statement 1 is sufficient as this is giving us the case of when an event is not happening ...question asked is when the event happens atleast once i.e 1-(when the event does not happens).
Statement 2 is sufficient cause as we know the probability of event happening so we can find when atleast once.

Bunuel niks18 VeritasPrepKarishma



For Statement 2:
How can we link outcome of event A with independent events in question stem?
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Though it is long time, let me try to explain how 2nd one is giving ans. 1st one is obvious.

let say there are 3 balls in a pot A, B and C. and you need to select one(A) randomly. when it is asked at least one, that means you can be successful 9 times to any 1 time. but don't fail all 9 times. means One attempted 9 times and all those 9 times one selected B or C. So if we find the probability to fail every time. We can find the probability of at least one time. probability to fail one time is 2/3. so for failing 9 times will be \((2/3)^9\).

pass at least one time will be 1 - \((2/3)^9\).

Hope That helps.

Hi guys,

Just wondering, if we had to solve ST2, how would we approach it using COMBINATORICS?

My thoughts:

P(A in 1 trial) = 1/3 => A=1, A'=2, Total Outcomes = A + A' = 3

Outcomes = 3
Trials = 9
# of Possible Outcomes = 3^9

Choose A' in 1st try * Choose A' in 2nd try* ... * Choose A' in 9th try = (2C1)^9 = 2^9

P(A in at least 1 of 9) = 1 - P(A' in all 9) = 1 - 2^9/3^9

Is this correct?

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