Difficult Math Question #9

P(R)*P(W) * 2

4/27

Soln:
3 red, 4 black and 2 white

total 9 balls

the first one can be either a red or a white
thus p(R1) = 3/9
and p(W1) = 2/9

since the balls are replaced, the second one can be either a white or red
thus p(W2) = 2/9
and p(R2) = 3/9

thus probability to draw either a red or white ball
= p(R1)*p(W2) + p(W1)*p(R2)
= 3/9*2/9 + 2/9*3/9
= 2(2/27)
= 4/27

Ans: D

800guy wrote:i'll post the OA when a few people have responded...

A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?

(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

(probability of a white and a red, whatever the order):

Hence
P(take a red and then a white) = 3/9 * 2/9 = 2/27
P(take a white and then a red) = 2/9 * 3/9 = 2/27
=> Total P =4/27, answer D.

IMHO, not a very difficult probability problem... I've seen much worse

OA:

Case I: Red ball first and then white ball

P1 = 3/9*2/9= 2/27

Case 2: White ball first and then red ball

P2 = 2/9*3/9 = 2/27

Therefore total probability: p1 + p2 = 4/27

A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?

(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

3R, 4B, 2W balls. Total = 9 balls
prob(RW) = Prob(R)*Prob(W) (independent events as the ball is put back after drawn)
= (3/9)*(2/9) = 2/27
Also the case when White ball is drawn first and Red ball next should also be considered.
prob(WR) = Prob(W)*Prob(R) = 2/27
prob(drawing red and white ball in 2 successive draws) = 2/27 + 2/27 = 4/27
IMO D