Q. 3. A bag contains 5 white and 3 black balls.
Two balls are drawn at random without replacement.
Determine the probability of getting both the balls
black.
Solution. We have to get 2 black balls from
3 black balls.
Favourable cases = 3C2
Total number of cases = 8C2
| |
3C2
|
| Required probability = |
|
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8C2
|
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(3!)/[(2!) X (3 - 2)!]
|
| or Probability = |
|
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(8!)/[(2!) X (8 - 2)!]
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or Probability = 3/28
Q. 4. A jar contains 7 red balls, 5 green balls,
4 blue balls, and 3 white balls. A sample of size
7 balls is selected at random without replacement.
Find the probability that the sample contains 2 red
balls, 2 green balls, 2 blue balls, and 1 white ball.
(June 2002)
Solution. Total number of balls = 7 + 5 +
4 + 3 = 19
Favourable cases = 7C2 X 5C2
X 4C2 X 3C1
A sample of size 7 balls is selected at random without
replacement.
Therefore, total number of cases = 19C7
| |
7C2 X 5C2
X 4C2 X 3C1
|
| Required probability = |
|
| |
19C7
|
| |
{(7!)/[(2!)(5!)]} X {(5!)/[(2!)(3!)]}
X {(4!)/[(2!)(2!)]} X {(3!)/[(1!)(2!)]}
|
| or Probability = |
|
| |
{(19!)/[(7!)(12!)]}
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or Probability = 0.075
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