Abhi077
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