Question

Let X_1 and X_2 be a random sample from a Poisson distribution with ? = 1. Let Y = min(X_1, X_2). The value of P[Y = 1] is (A) (2e-3)/e^2 (B) (2e-1)/e^2 (C) (e-1)/e (D) 1/e Recall that the probability mass function of Poisson random variable X with parameter ? is f(x) = P(X = x) = (e^(-?) * ?^x) / x! , x = 0, 1, ...

          Let X_1 and X_2 be a random sample from a Poisson distribution with ? = 1. Let Y = min(X_1, X_2). The value of P[Y = 1] is

(A) (2e-3)/e^2
(B) (2e-1)/e^2
(C) (e-1)/e
(D) 1/e

Recall that the probability mass function of Poisson random variable X with parameter ? is

f(x) = P(X = x) = (e^(-?) * ?^x) / x! , x = 0, 1, ...

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Let X1 and X2 be a random sample from a Poisson distribution with ? = 1. Let Y = min(X1, X2). The value of P[Y = 1] is

(A) (2e-3)/e^2
(B) (2e-1)/e^2
(C) (e-1)/e
(D) 1/e

Recall that the probability mass function of Poisson random variable X with parameter ? is

f(x) = P(X = x) = (e^(-?) * ?^x) / x! , x = 0, 1, ...

Added by Amparo S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

09/06/2023

Step 1

This means that one of them is 1 and the other is greater than or equal to 1. We can write this as: $P(Y=1) = P(X_1=1, X_2 \geq 1) + P(X_1 \geq 1, X_2=1) - P(X_1=1, X_2=1)$ Now, we can use the probability mass function of the Poisson distribution to find these Show more…

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Let X1 and X2 be a random sample from a Poisson distribution with λ. Let Y = min(X1, X2). The value of P(Y = 1) is (A) (B) (C) Recall that the probability mass function of a Poisson random variable with parameter λ is f(r) = (e^(-λ) * λ^r) / r!

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