Bag 1 contains 3 red and 4 black balls and bag 2 contains 4red and 5 black balls. two balls are transferred at random from bag 1 to 2 and then a ball is drawn from bag2. The ball so drawn is found to be red in colour. Find the probability that the transferred balls were both black.
Answers
Answer:
Probability : 11/77
Step-by-step explanation:
Bag1: 3 R + 4 B Bag2 : 4 R + 5 B
2 Balls transferred from bag1 to bag 2.
1) probability that 1st one is black followed by 2nd ball is black:
p1 = 4/(4+3) * 3/(3+3) = 2/7
Now bag2 contains: 4 R + 7 B
Probability that a ball drawn from bag2 is red: p2 = 4/(4+7) = 4/11
2) Probability that 1st is black & 2nd is red:
p3: 4/7 * 3/6 = 2/7
Now bag2 has 5 R + 6 B
Probability that ball drawn from bag2 is red : p4 = 5/(5+6) = 5/11
3) Probability that 1st is red & 2nd is Black:
p5: 3/7 * 4/6 = 2/7
Now bag2 has : 5 R + 6 B
Probability that ball from bag2 is red: p6: 5/11
4) Probability that 1st is red and 2nd is red :
p7: 3/7 * 2/6 = 1/7
Now bag2 has : 6 R + 5 B
Probability that ball drawn from bag2 is red: p8 = 6/11
Overall Probability that the ball drawn from bag2 is red:
p9 = 2/7 * 4/11 + 2/7 * 5/11 + 2/7 * 5/11 + 1/7 * 6/11
= 34/77
Now given that the ball drawn is red in color, the probability that the transferred balls were both black = p1 / p9
= 2/7 / (34/77) = 11/17
Answer:
Step-by-step explanation:
given that the ball drawn is red in color, the probability that the transferred balls were both black = p1 / p9
= 2/7 / (34/77) = 11/17