Answer to Question #148347 in Statistics and Probability for testans

Question

Bag A consists of 40 balls, 20% of them are red, the rest of them are green. Bag B consists of 30 balls, 70% of them are red, the rest of them are green. A ball is taken at random from Bag A and put into Bag B. Then, a ball is then taken at random from Bag B. Find the probability of having a red ball eventually.

Answer 1

Bag A consists of 20% red color balls = 0.2*40 = 8 balls

Remaining 80% ball are green color = 0.8*40 = 32 balls

Bag B consists of 70% red color balls = 0.7*30 = 21 balls

Remaining 30% ball are green color = 0.3*30 = 9 balls

Probability of getting a red ball from bag B = Transfer a red ball from bag A & select a red ball from bag B or Transfer a green ball from bag A and select a red ball from bag B

$= \frac {8C1*22C1}{40C1*31C1}+\frac {32C1*21C1}{40C1*31C1}$

$=\frac {8*22}{40*31}+\frac {32*21}{40*31}=0.1419+0.5419=0.68387$