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प्रश्न
The probability distribution of a random variable X is given below:
| X | 0 | 1 | 2 | 3 |
| P(X) | k | `"k"/2` | `"k"/4` | `"k"/8` |
Find P(X ≤ 2) + P (X > 2)
उत्तर
P(X ≤ 2) + P (X > 2) = `14/15 + 1/15`
= `(14 + 1)/15`
= `15/15`
= 1.
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Solution:
Here, n = 4
p = probability of defective device = 10% = `10/100 = square`
∴ q = 1 - p = 1 - 0.1 = `square`
X ∼ B(4, 0.1)
`P(X=x)=""^n"C"_x p^x q^(n-x)= ""^4"C"_x (0.1)^x (0.9)^(4 - x)`
P[At most one defective device] = P[X ≤ 1]
= P[X=0] + P[X=1]
= `square+square`
∴ P[X ≤ 1] = `square`
