Abhi077 wrote:

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 \(\frac{3}{7}\)

2) The number of left-handed children is less than the number of right handed children in the class

Target question: How many of the children are left-handed? Given: There are 15 children in a gym class and 2 children are to be randomly chosen to perform a demonstration 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-handedSince 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 classThere 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

Answer: A

Cheers,

Brent

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