Let X be a discrete random variable with possible values {0, 1, 2, ...} and the following probability mass function: P(X=0)=4/5 and for k€{1, 2, 3, ...} P(X=k)=1/10*(2/3)^k.a. Verify that the above is a probability mass function.b. For k€{1, 2, ...}, find P(X>k|X>1). Exercise 2.49. Let X be a discrete random variable with possible values{0, 1, 2, . } and the following probability mass function: P(X = 0) = andfor k e {1, 2, 3,...}%3D%3DkollyP(X = k) = · (G).%3D(a) Verify that the above is a probability mass function.(b) For ke {1, 2, . }, find P(X > k| X > 1).

We have been given that, P(X=0) = 4/5 and for k ∈ {1, 2, 3...} we have, P(X=k)=11023k Part a: To…

Answered: Let X be a discrete random variable…

Algebra & Trigonometry with Analytic Geometry

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ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.8: Probability

Problem 20E

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Question

Let X be a discrete random variable with possible values {0, 1, 2, ...} and the following probability mass function: P(X=0)=4/5 and for k€{1, 2, 3, ...} P(X=k)=1/10*(2/3)^k.

a. Verify that the above is a probability mass function.

b. For k€{1, 2, ...}, find P(X>k|X>1).

Exercise 2.49. Let X be a discrete random variable with possible values
{0, 1, 2, . } and the following probability mass function: P(X = 0) = and
for k e {1, 2, 3,...}
%3D
%3D
k
olly
P(X = k) = · (G).
%3D
(a) Verify that the above is a probability mass function.
(b) For ke {1, 2, . }, find P(X > k| X > 1).

Expert Solution

Step 1

We have been given that,

P(X=0) = 4/5 and for k ∈ {1, 2, 3...} we have,

$P (X = k) = \frac{1}{10} {(\frac{2}{3})}^{k}$

Part a:

To show that the given mass function is a probability mass function we need to sow that total probability is equal to 1.

Thus, we have,

$T o t a l p r o b a b i l i t y = \sum_{k = 0}^{\infty} P (X = k) = \frac{4}{5} + \sum_{k = 1}^{\infty} P (X = k) = \frac{4}{5} + \frac{1}{10} \sum_{k = 1}^{\infty} {(\frac{2}{3})}^{k} = \frac{4}{5} + \frac{1}{10} (\frac{2}{3} + {(\frac{2}{3})}^{2} + {(\frac{2}{3})}^{3} + {(\frac{2}{3})}^{4} + . . . . .) = \frac{4}{5} + \frac{1}{10} \times \frac{2}{3} (1 + \frac{2}{3} + {(\frac{2}{3})}^{2} + {(\frac{2}{3})}^{3} + {(\frac{2}{3})}^{4} + . . . .) = \frac{4}{5} + (\frac{1}{10} \times \frac{2}{3}) {(1 - \frac{2}{3})}^{- 1} = \frac{4}{5} + (\frac{1}{10} \times \frac{2}{3}) {(\frac{1}{3})}^{- 1} = \frac{4}{5} + \frac{1}{5} = 1$

Thus, we have total probability = 1.

Hence, above given mass function is a probability mass function.

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