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# If there are 15 children in a gym class and 2 children are to be randomly chosen to perform a demonstration, how many of the children are left-handed?

Difficulty: Sub-600 Level, Probability, Data Sufficiency (DS)

Monday 25 July 2022,

If there are 15 children in a gym class and 2 children are to be randomly chosen to perform a demonstration, how many of the children are left-handed?
1) The probability of no left-handed children being chosen is 3/7
2) The number of left-handed children is less than the number of right handed children in the class

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EXPLANATION:
Statement 1: The probability of no left-handed children being chosen is 3/7
P(no left-handers among those selected) = P(both right-handers among those selected)
So, P(both right-handers among those selected) = 3/7
In other words, P(1st selection is right-handed AND 2nd selection is right-handed) = 3/7
So, P(1st selection is right-handed) x P(2nd selection is right-handed) = 3/7

Let R = # of right-handers in the group

We can write: R/15 x (R - 1)/14 = 3/7
Simplify to get: (R² - R)/210 = 3/7
Cross multiply to get: 7(R² - R) = 3(210)
Divide both sides by 7 to get: R² - R = 3(30)
Or....: R² - R = 90
Then: R² - R - 90 = 0
Factor: (R - 10)(R + 9) = 0
So, EITHER R = 10 OR R = -9
Since R cannot be negative, we know that R = 10 (i.e., there are 10 right-handers)

If 10 children are right-handed, then 5 children are left-handed
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The number of left-handed children is less than the number of right handed children in the class
There are several scenarios that satisfy statement 2. Here are two:
Case a: There are 4 left-handers and 11 right-handers. In this case, the answer to the target question is there are 4 left-handers
Case b: There are 5 left-handers and 10 right-handers. In this case, the answer to the target question is there are 5 left-handers
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT