92. Bag I contains 3 Red and 4 Black balls and B II contains 4 Red and % Black balls. One ball is transferred from Bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be Red in colour. Find the probability that the transferred ball is Black.
92. Bag I contains 3 Red and 4 Black balls and B II contains 4 Red and % Black balls. One ball is transferred from Bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be Red in colour. Find the probability that the transferred ball is Black.
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1 Answer
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92. Let E1 = Ball transferred from Bag I to Bag II is red
E2 = Ball transferred from Bag I to Bag Ii is black
A = Ball drawn from Bag II is red in colour
P (E1) =3/7 and P (E2) = 4/7
Let A be the event that the ball drawn is red.
When a red ball is transferred from bag I to II,
P (A | E1) = 5/10 = 1/2
When a black ball is transferred from bag I to II,
P (A | E2) = 4/10 = 2/5
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P (2 obtained on even numbered toss) = k (let)
P (2) =
P (
If x = 0, y = 6, 7, 8, 9, 10
If x = 1, y = 7, 8, 9, 10
If x = 2, y = 8, 9, 10
If x = 3, y = 9, 10
If x = 4, y = 10
If x = 5, y = no possible value
Total possible ways = (5 + 4 + 3 + 2 + 1) * 2
= 30
Required probability
P (2W and 2B) = P (2B, 6W) × P (2W and 2B)
+ P (3B, 5W) × P (2W and 2B)
+ P (4B, 4W) × P (2W and 2B)
+ P (5B, 3W) × P (2W and 2B)
+ P (6B, 2W) × P (2W and 2B)
(15 + 30 + 36 + 30 + 15)
Let probability of tail is
⇒ Probability of getting head =
∴ Probability of getting 2 heads and 1 tail
ax2 + bx + c = 0
D = b2 – 4ac
D = 0
b2 – 4ac = 0
b2 = 4ac
(i) AC = 1, b = 2 (1, 2, 1) is one way
(ii) AC = 4, b = 4
(iii) AC = 9, b = 6, a = 3, c = 3 is one way
1 + 3 + 1 = 5 way
Required probability =
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