A bag contains 3 red, 2 white and 2 black balls. Two balls are drawn at random and none of them is found to be a white ball. The probability that both balls are red is (a/b) (where a,b are coprime) then b-a is equal to

Question

A bag contains 3 red, 2 white and 2 black balls. Two balls are drawn at random and none of them is found to be a white ball. The probability that both balls are red is (a/b) (where a,b are coprime) then b-a is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

E=2 balls drawn, none of them is white E1= both balls red E2= both balls black E3= one red, one black P(E1 / E)=(P (E / E1) ⋅ P (E1)/P (E / E1) ⋅ P (E1) + P (E / E2) ⋅ P (E2) + P (E / E3) ⋅ P (E3)) =(( 3C2/ 7C2)/( 3C2) 7C2 + ( 2C2) 7C2 + ( 3C2 ⋅ 2C1/ 7C2)=( 3C2/ 3C2 + 2C2 + 3C1 ⋅ 2C1) =(3/3 + 1 + 3 . 2)=(3/10)

Q. A bag contains 3 red, 2 white and 2 black balls. Two balls are drawn at random and none of them is found to be a white ball. The probability that both balls are red is ba​ (where a,b are coprime) then b−a is equal to

Q. A bag contains $3$ red, $2$ white and $2$ black balls. Two balls are drawn at random and none of them is found to be a white ball. The probability that both balls are red is $\frac{a}{b}$ (where $a, b$ are coprime) then $b - a$ is equal to