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Question
For the following probability density function (p. d. f) of X, find P(X < 1) and P(|x| < 1)
`f(x) = x^2/18, -3 < x < 3`
= 0, otherwise
Solution
We have
P(X < 1) = `int_-3^1 f(x) dx`
= `int_-3^1 x^2/18dx`
= `[x^3/54]_-3^1`
= `1/54 - ((-27)/54)`
= `28/54`
= `14/27`
= 0.5185
Now |X| < 1
`\implies` ± X < 1
∴ X < 1, – X < 1,
i.e. X > – 1
i.e. – 1 < X < 1
∴ Required Probability = `int_-1^1 f(x)dx`
= `int_-1^1 x^2/18dx`
= `[x^3/54]_-1^1`
= `1/54 + 1/54`
= `2/54`
= `1/27`
= 0.03704
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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`
