Bunuel
A bag contains blue and red balls only.
If one ball is drawn, the probability of selecting a blue ball is 1/2.
If two balls are drawn randomly without replacement, the probability of selecting a blue and then a red ball is 5/18.
Two balls are drawn randomly without replacement, what is the probability of drawing a red ball, given that the first ball drawn was blue?
(A) 2/9
(B) 5/18
(C) 4/9
(D) 1/2
(E) 5/9
Number of blue balls = b
Number of red balls = r
If one ball is drawn, the probability of selecting a blue ball is 1/2.\(\frac{b}{b+r} = \frac{1}{2}\)
Upon solving we get
\(b = r\) ---------- (1)
If two balls are drawn randomly without replacement, the probability of selecting a blue and then a red ball is 5/18.\(\frac{b}{b+r} * \frac{r}{b+r-1} = \frac{5}{18}\)
As b = r, let's replace the same
\(\frac{b}{2b} * \frac{b}{2b-1} = \frac{5}{18}\)
\(\frac{1}{2} * \frac{b}{2b-1} = \frac{5}{18}\)
18b = 20b - 10
b = 5
b = r = 5
Two balls are drawn randomly without replacement, what is the probability of drawing a red ball, given that the first ball drawn was blue?Number of red balls = 5
Total number of balls remaining = 9
Required probability = 5/9
Option E