Let-the-independent-random-variables-X-1-and-X-2-have-binomial-distribution-with-parameters-n-1-3-p-2-3-and-n-2-4-p-1-2-respectively-Compute-P-X-1-X-2-

Question Number 149463 by jlewis last updated on 05/Aug/21

Let the independent random variables

X_{1} and X_{2} have binomial distribution

with parameters n_{1} = 3, p = 2 / 3 and n_{2} = 4

p = 1 / 2 respectively .

Compute P (X_{1} = X_{2})

Answered by Olaf_Thorendsen last updated on 05/Aug/21

∙ P (X_{1} = k) = C_{k}^{n_{1}} p_{1}^{k} {(1 - p_{1})}^{n_{1} - k}

P (X_{1} = k) = C_{k}^{3} {(\frac{2}{3})}^{k} {(\frac{1}{3})}^{3 - k} = \frac{1}{27} C_{k}^{3} 2^{k}

∙ P (X_{2} = k) = C_{k}^{n_{2}} p_{2}^{k} {(1 - p_{2})}^{n_{2} - k}

P (X_{2} = k) = C_{k}^{4} {(\frac{1}{2})}^{k} {(\frac{1}{2})}^{4 - k} = \frac{1}{16} C_{k}^{4}

∙ P (X_{1} = X_{2}) = P (X_{1} = 0) P (X_{2} = 0)

+ P (X_{1} = 1) P (X_{2} = 1)

+ P (X_{1} = 2) P (X_{2} = 2)

+ P (X_{1} = 3) P (X_{2} = 3)

= \frac{C_{0}^{3} 2^{0} C_{0}^{4} + C_{1}^{3} 2^{1} C_{1}^{4} + C_{2}^{3} 2^{2} C_{2}^{4} + C_{3}^{3} 2^{3} C_{3}^{4}}{27.16}

= \frac{1 + 24 + 72 + 32}{432} = \frac{43}{144} \approx 29, 9 % .