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A bag contains 5 red and 7 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively of different colours?
- a)7/99
- b)11/99
- c)14/99
- d)19/99
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A bag contains 5 red and 7 white balls. Four balls are drawn out one b...
Balls are picked in two manners – RWRW or WRWR
So probability = (5/12)*(7/11)*(4/10)*(6/9) + (7/12)*(5/11)*(6/10)*(4/9) = 14/99
So probability = (5/12)*(7/11)*(4/10)*(6/9) + (7/12)*(5/11)*(6/10)*(4/9) = 14/99
Most Upvoted Answer
A bag contains 5 red and 7 white balls. Four balls are drawn out one b...
Problem: A bag contains 5 red and 7 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternatively of different colours?
Solution:
Total number of ways of drawing 4 balls out of 12 = $\binom{12}{4}$
Let’s consider the first ball drawn. It can either be red or white. Without loss of generality, let’s assume it is red. Then the second ball drawn must be white. The third ball must again be red, and the fourth ball must be white. The total number of ways of drawing the balls in this manner is:
Number of ways of drawing 1 red ball out of 5 $\times$ Number of ways of drawing 1 white ball out of 7 $\times$ Number of ways of drawing 1 red ball out of 4 $\times$ Number of ways of drawing 1 white ball out of 6
= 5 $\times$ 7 $\times$ 4 $\times$ 6 = 840
Similarly, if we assume that the first ball drawn is white, then the total number of ways of drawing the balls in this manner is:
Number of ways of drawing 1 white ball out of 7 $\times$ Number of ways of drawing 1 red ball out of 5 $\times$ Number of ways of drawing 1 white ball out of 4 $\times$ Number of ways of drawing 1 red ball out of 3
= 7 $\times$ 5 $\times$ 4 $\times$ 3 = 420
Therefore, the total number of ways of drawing the balls alternatively of different colours is 840 + 420 = 1260.
Hence, the probability of drawing the balls alternatively of different colours is:
$\frac{\text{Number of ways of drawing the balls alternatively of different colours}}{\text{Total number of ways of drawing 4 balls out of 12}}$
$= \frac{1260}{\binom{12}{4}}$
$= \frac{1260}{495}$
$= \frac{28}{99}$
Therefore, the answer is (C) 14/99.
Solution:
Total number of ways of drawing 4 balls out of 12 = $\binom{12}{4}$
Let’s consider the first ball drawn. It can either be red or white. Without loss of generality, let’s assume it is red. Then the second ball drawn must be white. The third ball must again be red, and the fourth ball must be white. The total number of ways of drawing the balls in this manner is:
Number of ways of drawing 1 red ball out of 5 $\times$ Number of ways of drawing 1 white ball out of 7 $\times$ Number of ways of drawing 1 red ball out of 4 $\times$ Number of ways of drawing 1 white ball out of 6
= 5 $\times$ 7 $\times$ 4 $\times$ 6 = 840
Similarly, if we assume that the first ball drawn is white, then the total number of ways of drawing the balls in this manner is:
Number of ways of drawing 1 white ball out of 7 $\times$ Number of ways of drawing 1 red ball out of 5 $\times$ Number of ways of drawing 1 white ball out of 4 $\times$ Number of ways of drawing 1 red ball out of 3
= 7 $\times$ 5 $\times$ 4 $\times$ 3 = 420
Therefore, the total number of ways of drawing the balls alternatively of different colours is 840 + 420 = 1260.
Hence, the probability of drawing the balls alternatively of different colours is:
$\frac{\text{Number of ways of drawing the balls alternatively of different colours}}{\text{Total number of ways of drawing 4 balls out of 12}}$
$= \frac{1260}{\binom{12}{4}}$
$= \frac{1260}{495}$
$= \frac{28}{99}$
Therefore, the answer is (C) 14/99.
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