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प्रश्न
A bag contains 8 red, 3 white and 9 blue balls. If three balls are drawn at random, determine the probability that all the balls are of different colours.
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उत्तर
Out of 20 balls, three balls can be drawn in 20C3 ways.
∴ Total number of elementary events = 20C3
Out of eight red balls, one red ball can be drawn in 8C1 ways.
Out of three white balls, one white ball can be drawn in 3C1.
Out of nine blue balls, one blue ball can be drawn in 9C1 ways.
So, favourable number of elementary events = 8C1 × 3C1 × 9C1
Hence, required probability =\[\frac{^{8}{}{C}_1 \times ^{3}{}{C}_1 \times^{9}{}{C}_1}{^{20}{}{C}_3} = \frac{8 \times 3 \times 9}{60 \times 19} = \frac{18}{95}\]
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संबंधित प्रश्न
Which of the following can not be valid assignment of probabilities for outcomes of sample space S = {ω1, ω2,ω3,ω4,ω5,ω6,ω7}
| Assignment | ω1 | ω2 | ω3 | ω4 | ω5 | ω6 | ω7 |
| (a) | 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
| (b) | `1/7` | `1/7` | `1/7` | `1/7` | `1/7` | `1/7` | `1/7` |
| (c) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 |
| (d) | –0.1 | 0.2 | 0.3 | 0.4 | -0.2 | 0.1 | 0.3 |
| (e) | `1/14` | `2/14` | `3/14` | `4/14` | `5/14` | `6/14` | `15/14` |
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| Elementary events: | w1 | w2 | w3 | w4 | w5 | w6 | w7 |
| (iv) |
\[\frac{1}{14}\]
|
\[\frac{2}{14}\]
|
\[\frac{3}{14}\]
|
\[\frac{4}{14}\]
|
\[\frac{5}{14}\]
|
\[\frac{6}{14}\]
|
\[\frac{15}{14}\]
|
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