A box contains red and blue balls only. If there are 8 balls in total

Suppose there are x red balls, and thus, 8-x blue balls.
1) P(both red w/o rep) = x/8 *(x-1)/7 = 5/14. since quad, we'll get 2 values of x, but the no of balls has to be positive. So we take the +ve one. SUFF
2) P(1st red, 2nd blue w/o rep) = x/8 * (8-x)/7 = 15/56 => x*(8-x) = 15. This gives 2 values of x : 5,3. Thus not SUFF, since red balls can be either 5 or 3.

Ans A.

I went with A


1) Sufficient

x/8 * x/7 = 5/14 <--- solvable and gives ratio. I don't do the actual math, I just know that it's possible....

2) Insufficient
Since the 15/56 results from p(blue)*p(red) there is no way to determine one of them alone from the information given..

number of red balls (R), number of Blue balls (B)
1) probability that two balls randomly selected without replacement, are red = \((R/8) * ({R-1}/7) = 5/14\)
we get R(R-1) = 20
R = 5,-4
R cannot be negative so number of red balls (R) = 5

Sufficient Alone.

2) probability that two balls randomly selected without replacement, the first ball is red and the second ball is blue = \((R/8) * (B/7) = 15/56\)
we get, R*B = 15
we know that R+B = 8
So only possible values are 3 and 5 but which one is red is unclear.
Insufficient Alone.

Answer:A

R+B=8; R=?
1. 5/14=getting two R w/o replacement=>R(R-1)/8*7=> as R can be only positive so only one answer; Sufficient
2. 15/56=one R one B=>R(8-R)/8*7=> R has two positive values; Not sufficient
Hence answer is A

Thanks,

For this type of question we should only know the principles of probability and don't need to make any calculations to find the correct answer:

We know that this box contain only red and blue balls (important: this approach will be work only for two types of items).
So if we have information about probability randomly selected balls of ONE color (any quantity) we can find how much balls of each color in box.

1) in this statement we have information about probability of two balls the SAME color - this is sufficient and we don't need any calculations, because this is DS task and we should save time

If we have probability for balls with DIFFERENT colors we can't find how much balls of each color in box
2) in this statement we have information about probability of two balls the same color - this is insufficient and we don't need any calculations, because this is DS task and we should save time

So answer is A

MAGOOSH OFFICIAL SOLUTION:

Hi All,
WRT official soln

is it not (RB / 56 ) = 1/2 * ( 15/56) because there are two ways ...

1. Non red & then red
2. Red and then non red.


Kindly clarify

Hi

Second statement clearly specifies that the probability of selecting first ball as red and second ball as blue is = 15/56. So its only the second case:- out of the two cases specified by you.

Bunuel in second statement, I did rC1 * 8-rC1/ 8C2 = 15/56 but it did not work, can we not use the combinations formula here?