Question

What is the probability distribution of X when X~B(1,1/25)?

P(X= 0) = 0.96
P(X= 1) = 0.04

P(X= 0) = 0.75
P(X= 1) = 0.25

P(X= 0) = 0.04
P(X= 1) = 0.96

P(X= 0) = 0.25
P(X= 1) = 0.75

Answer 1

The probability distribution of X when X~B(1,1/25) is (Option A)

• P(X= 0) = 0.96
• P(X= 1) = 0.04

See the explanation below.

What is the explanation to the above solution?

Given X~B(n,p)

P(x=1) = C¹ₙ * P¹ * (1-P)ⁿ⁻¹ (n≥1)

Thus, X~B [1, 1/25]

1/25 = 0.04

hence, p(x=0)

= C⁰₁ * 0.04⁰ (1 - 0.04)¹

= 0.96

P (x=1)

=  C¹₁ 0.04¹ (1-0.04)⁰

= 0.04

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