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by rtaha2412 » Mon Jan 03, 2011 3:30 pm
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?

a 0.343
b 0.147
c 0.189
d 0.063
e 0.027
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by anshumishra » Mon Jan 03, 2011 3:34 pm
rtaha2412 wrote:The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?

a 0.343
b 0.147
c 0.189
d 0.063
e 0.027
3C2*(0.3)^2*(0.7) = 0.189

C
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by GMATGuruNY » Mon Jan 03, 2011 4:06 pm
rtaha2412 wrote:The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?

a 0.343
b 0.147
c 0.189
d 0.063
e 0.027
P(1st visitor buys candy) = 3/10
P(2nd visitor buys candy) = 3/10
P(3rd visitor doesn't buy candy) = 7/10

Since we need all of the events above to happen together, we multiply the fractions: 3/10 * 3/10 * 7/10 = 63/1000.

Since the visitor who doesn't buy candy could be 1st, 2nd, or 3rd, we multiply the result above by 3:

3 * 63/1000 = 189/1000.

The correct answer is C.
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by Jeff@TargetTestPrep » Tue Jan 02, 2018 11:15 am
GMATGuruNY wrote:
rtaha2412 wrote:The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two visitors will buy a pack of candy?

a 0.343
b 0.147
c 0.189
d 0.063
e 0.027


We need to determine the probability that two out of three visitors will buy a pack of candy:

P(Y-Y-N) = 0.3 x 0.3 x 0.7 = 0.063

Since there are 3 ways -- (Y-Y-N), (Y-N-Y), or (N-Y-Y) -- in which two of the three visitors can buy a pack of candy, the overall probability is 3 x 0.063 = 0.189.

Answer: C

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