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Source Grockit: DS: Probability

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by polter » Wed Apr 25, 2012 6:38 pm
An ice cream truck is passing out free ice cream to each child at a playground. The driver randomly hands out only chocolate, vanilla and strawberry ice cream. What is the probability that a certain child will receive either chocolate or strawberry ice cream?

(1) The probability that a certain child will receive chocolate ice cream is 1/10.

(2) The probability that a certain child will receive vanilla is 1/4.

[spoiler]B[/spoiler]

I think since he is giving out randomly, then each will be equally likely event. Hence prob of chocolate = prob of vanilla = prob of strawberry.
Is this line of thinking incorrect?
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Source: — Data Sufficiency |
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by shantanu86 » Wed Apr 25, 2012 7:35 pm
polter wrote:An ice cream truck is passing out free ice cream to each child at a playground. The driver randomly hands out only chocolate, vanilla and strawberry ice cream. What is the probability that a certain child will receive either chocolate or strawberry ice cream?

(1) The probability that a certain child will receive chocolate ice cream is 1/10.

(2) The probability that a certain child will receive vanilla is 1/4.

B

I think since he is giving out randomly, then each will be equally likely event. Hence prob of chocolate = prob of vanilla = prob of strawberry.
Is this line of thinking incorrect?
Hi polter,

Here the correct answer is [spoiler][/spoiler].

The driver has only three kinds of ice-cream and every single child gets one type.
Therefore,
P(Child gets ice-cream) = 1.

Now, the probability of getting a particular type can depend on a number of factors.. like quantity of each type present with the driver.

P(vanilla/ice-cream) = 1/4 .. indicates vanilla stock is 1/4th of total and rest is chocolate+vanilla.

Therefore,

P(reqd) = P(Child gets ice-cream)X (1-P(vanilla/ice-cream))
p(reqd) = 3/4

Hence, (2) alone is sufficient=> [spoiler] [/spoiler]is the correct answer.

Hope it helps!!
If you feel like it, hit thanks :)
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