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प्रश्न
Solve the following problem:
Following is the probability distribution of a r.v.X.
| X | – 3 | – 2 | –1 | 0 | 1 | 2 | 3 |
| P(X = x) | 0.05 | 0.1 | 0.15 | 0.20 | 0.25 | 0.15 | 0.1 |
Find the probability that X is odd.
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उत्तर
P(X is odd)
= P(X = –3 or X = –1 or X = 1 or X = 3)
= P(X = –3) + P(X = – 1) + P(X = 1) + P(X = 3)
= 0.05 + 0.15 + 0.25 + 0.10
= 0.55
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Solution:
Here, n = 4
p = probability of defective device = 10% = `10/100 = square`
∴ q = 1 - p = 1 - 0.1 = `square`
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∴ P[X ≤ 1] = `square`
