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Fifteen candies are to be divided up among six children so

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Fifteen candies are to be divided up among six children so [#permalink] New post   20 Jul 2020, 05:08
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Fifteen candies are to be divided up among six children so that each child receives at least two candies.


Quantity A
Quantity B
The probability that any child receives six candies
0



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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Re: Fifteen candies are to be divided up among six children so [#permalink] New post   20 Jul 2020, 05:08
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Re: Fifteen candies are to be divided up among six children so [#permalink] New post   27 Jul 2020, 23:34
Six children receive atleast 2 candles. 6*2 = 12 candles.

We are left with 15-12 = 3 candles. If all 3 are given to ONE child, the child still has max 5 candles.

Hence, probability of a child getting 6 candles = 0. C.
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Re: Fifteen candies are to be divided up among six children so [#permalink] New post   28 Jul 2020, 01:09
"each child receives at least two candies." So this means that at least 10 candies should be distributed amongst five children. This leaves five candies at best, and hence we get a perfect 0 as probability since the case of someone getting six candies is impossible.
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Re: Fifteen candies are to be divided up among six children so [#permalink] New post   28 Jul 2020, 02:25
No child can receive 6 candies. Because each one has to get 2 candies minimum. There are 6 children and 15 candies. 6*2 = 12. 15-12=3. The maximum any child could get is 2+3=5 and not 6. So Probability (quantity A) is 0 and option C is the answer.
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Re: Fifteen candies are to be divided up among six children so [#permalink]
28 Jul 2020, 02:25
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