aditya.j wrote:Q) A bag contains 12 balls some of which are red and rest are blue. How many red balls are there in the bag?
1) The probability of randomly selecting two red balls from the bag is 1/11.
2) When two balls are selected randomly from the bag the probability of selecting one red and one blue ball is 9/22
I really couldn't figure this one out!
Let R = # red balls
Let B = # blue balls
Given: R+B=12
Target: find the value of R
Statement 1: P(2 red) = 1/11
P(2 red) = P(1st ball is red) x P(2nd ball is red)
= (R/12)(R-1/11)
= (R^2-R)/132
Since we're told that P(2 red) = 1/11, we can write: (R^2-R)/132 = 1/11
Multiply both sides by 132 to get R^2 - R = 12
Set equal to zero: R^2 - R - 12 = 0
Factor: (R-4)(R+3)=0
R = -3, 4
Since R cannot be negative, R must equal 4
SUFFICIENT
Statement 2: P(1 red and 1 blue) = 9/22
P(1 red and 1 blue) = P(red then blue
or blue then red)
= P(red then blue)
+ P(blue then red)
= (R/12)(B/11) + (B/12)(R/11)
= RB/66
Since we're told that P(1 red and 1 blue) = 9/22, we can write: RB/66 = 9/22
Multiply both sides by 66 to get RB=27
So, we know that RB=27 and R+B=12
case a: R=3 and B=9
case b: R=9 and B=3
Since R can have 2 different values, statement 2 is INSUFFICIENT
Answer =
A
Cheers,
Brent
