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A box contains 5 black and 4 white balls. A ball is drawn at random and its colour is noted. The ball is then put back in the box along with two additional balls of its opposite colour. If a ball is drawn again from the box, then the probability that the ball drawn now is black, is
- a)7/11
- b)5/11
- c)53/99
- d)48/99
Correct answer is option 'C'. Can you explain this answer?
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Solution:
Initial probability of drawing a black ball = P(B) = 5/9
Initial probability of drawing a white ball = P(W) = 4/9
After putting the ball back along with two additional balls of opposite colour, the total number of balls in the box becomes 9+2 = 11.
Now, there are two cases to consider:
Case 1: The first ball drawn is black
After putting the black ball back, the number of black balls in the box becomes 5+2 = 7 and the number of white balls becomes 4+1 = 5.
Probability of drawing a black ball in this case = P(B)×P(B|first ball is black)
= (5/9)×(7/11)
= 35/99
Case 2: The first ball drawn is white
After putting the white ball back, the number of black balls in the box becomes 5+1 = 6 and the number of white balls becomes 4+2 = 6.
Probability of drawing a black ball in this case = P(W)×P(B|first ball is white)
= (4/9)×(6/11)
= 24/99
Total probability of drawing a black ball in the second draw = Probability in case 1 + Probability in case 2
= (35/99) + (24/99)
= 59/99
Therefore, the probability that the ball drawn now is black is 53/99.
Answer: (c) 53/99
Initial probability of drawing a black ball = P(B) = 5/9
Initial probability of drawing a white ball = P(W) = 4/9
After putting the ball back along with two additional balls of opposite colour, the total number of balls in the box becomes 9+2 = 11.
Now, there are two cases to consider:
Case 1: The first ball drawn is black
After putting the black ball back, the number of black balls in the box becomes 5+2 = 7 and the number of white balls becomes 4+1 = 5.
Probability of drawing a black ball in this case = P(B)×P(B|first ball is black)
= (5/9)×(7/11)
= 35/99
Case 2: The first ball drawn is white
After putting the white ball back, the number of black balls in the box becomes 5+1 = 6 and the number of white balls becomes 4+2 = 6.
Probability of drawing a black ball in this case = P(W)×P(B|first ball is white)
= (4/9)×(6/11)
= 24/99
Total probability of drawing a black ball in the second draw = Probability in case 1 + Probability in case 2
= (35/99) + (24/99)
= 59/99
Therefore, the probability that the ball drawn now is black is 53/99.
Answer: (c) 53/99
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A box contains 5 black and 4 white balls. A ball is drawn at random and its colour is noted. The ball is then put back in the box along with two additional balls of its opposite colour. If a ball is drawn again from the box, then the probability that the ball drawn now is black, isa) 7/11b) 5/11c) 53/99d) 48/99Correct answer is option 'C'. Can you explain this answer?
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