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Find the Probability!!!

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by vidhya16 » Sat Jan 14, 2012 2:55 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?
* 0.343
* 0.147
* 0.189
* 0.063
* 0.027
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by mj78ind » Sat Jan 14, 2012 3:23 pm
vidhya16 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?
* 0.343
* 0.147
* 0.189
* 0.063
* 0.027
We can use Binomial here:
3C2(0.3)^2(.7)^1 = 3*(0.09)*(0.7) = 0.189

Hence C
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by GMATGuruNY » Sat Jan 14, 2012 9:18 pm
vidhya16 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?
* 0.343
* 0.147
* 0.189
* 0.063
* 0.027
P(exactly n) = P(one way) * total possible ways.

P(one way):
One way to get a good outcome is for the first visitor not to buy a pack of candy while the second and third visitors do.
P(1st visitor doesn't buy candy) = 7/10.
P(2nd visitor buys candy) = 3/10.
P(3rd visitor buys candy) = 3/10.
To determine the probability that all of these events happen together, we multiply the fractions:
7/10 * 3/10 * 3/10 = .063.

Total possible ways:
Since any of the 3 visitors could be the one not to buy candy, the result above must be multiplied by 3:
3 * .063 = .189.

The correct answer is C.
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by ArunangsuSahu » Sat Jan 14, 2012 10:11 pm
3C2*(0.3)^2*0.7=0.189
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by vidhya16 » Sun Jan 15, 2012 9:13 am
Thanks all.
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