Carcass wrote:
In a jar, 9 balls are white and the rest are red. If two balls are to be chosen at random from the jar without replacement, the probability that the balls chosen will both be white is 6/11. What is the number of balls in the jar?
(A) 10
(B) 11
(C) 12
(D) 13
(E) 15
P(both white) = P(1st chosen ball is white
AND 2nd chosen ball is white)
= P(1st chosen ball is white)
x P(2nd chosen ball is white)
Let R = the number of red balls in the jar
So, at the beginning of the experiment, the total number of balls = R + 9Continuing, we get....
P(both white) =
9/(R + 9) x 8/(R + 8)Aside: Before the first draw, there are R+9 balls, and 9 of them are white. This is why the 1st probability is 9/(R + 9)
After we draw a white ball on the first turn, there are R+8 balls remaining, and 8 of them are white. This is why the 2nd probability is 8(R+8)We are told that P(both white) = 6/11
So, we can write:
9/(R + 9) x 8/(R + 8) = 6/11
Simplify: 72/(R+9)(R+8) = 6/11
Divide both sides by 6 to get: 12/(R+9)(R+8) = 1/11
Cross multiply: (R+9)(R+8)(1) = (12)(11)
Simplify: R² + 17R + 72 = 132
Subtract 132 from both sides to get: R² + 17R - 60 = 0
Factor to get: (R + 20)(R - 3) = 0
Solve: R = -20 or R = 3
Since our must be a positive number, it must be the case that R = 3 (in other words there are 3 red balls)
What is the number of balls in the jar?Now that we know there are 3 red balls, we can add this to the original 9 white balls to get a total of 12 balls
Answer: C
Cheers,
Brent