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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 )

$$\mathrm{Let}\:\mathrm{the}\:\mathrm{independent}\:\mathrm{random}\:\mathrm{variables} \\ $$$$\mathrm{X}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{X}_{\mathrm{2}} \:\mathrm{have}\:\mathrm{binomial}\:\mathrm{distribution} \\ $$$$\mathrm{with}\:\mathrm{parameters}\:\mathrm{n}_{\mathrm{1}} =\mathrm{3},\mathrm{p}=\mathrm{2}/\mathrm{3}\:\mathrm{and}\:\mathrm{n}_{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{p}=\mathrm{1}/\mathrm{2}\:\:\mathrm{respectively}.\: \\ $$$$\mathrm{Compute}\:\mathrm{P}\left(\mathrm{X}_{\mathrm{1}} =\mathrm{X}_{\mathrm{2}} \right) \\ $$

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 ((2/3))^k ((1/3))^(3−k) = (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 ((1/2))^k ((1/2))^(4−k) = (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) = ((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)) = ((1+24+72+32)/(432)) = ((43)/(144)) ≈ 29,9%.

$$\bullet\:\mathrm{P}\left(\mathrm{X}_{\mathrm{1}} ={k}\right)\:=\:\mathrm{C}_{{k}} ^{{n}_{\mathrm{1}} } {p}_{\mathrm{1}} ^{{k}} \left(\mathrm{1}−{p}_{\mathrm{1}} \right)^{{n}_{\mathrm{1}} −{k}} \\ $$$$\mathrm{P}\left(\mathrm{X}_{\mathrm{1}} ={k}\right)\:=\:\mathrm{C}_{{k}} ^{\mathrm{3}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{k}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}−{k}} =\:\frac{\mathrm{1}}{\mathrm{27}}\mathrm{C}_{{k}} ^{\mathrm{3}} \mathrm{2}^{{k}} \\ $$$$\bullet\:\mathrm{P}\left(\mathrm{X}_{\mathrm{2}} ={k}\right)\:=\:\mathrm{C}_{{k}} ^{{n}_{\mathrm{2}} } {p}_{\mathrm{2}} ^{{k}} \left(\mathrm{1}−{p}_{\mathrm{2}} \right)^{{n}_{\mathrm{2}} −{k}} \\ $$$$\mathrm{P}\left(\mathrm{X}_{\mathrm{2}} ={k}\right)\:=\:\mathrm{C}_{{k}} ^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}−{k}} =\:\frac{\mathrm{1}}{\mathrm{16}}\mathrm{C}_{{k}} ^{\mathrm{4}} \\ $$$$\bullet\:\mathrm{P}\left(\mathrm{X}_{\mathrm{1}} =\mathrm{X}_{\mathrm{2}} \right)\:=\:\mathrm{P}\left(\mathrm{X}_{\mathrm{1}} =\mathrm{0}\right)\mathrm{P}\left(\mathrm{X}_{\mathrm{2}} =\mathrm{0}\right) \\ $$$$\:+\mathrm{P}\left(\mathrm{X}_{\mathrm{1}} =\mathrm{1}\right)\mathrm{P}\left(\mathrm{X}_{\mathrm{2}} =\mathrm{1}\right) \\ $$$$+\mathrm{P}\left(\mathrm{X}_{\mathrm{1}} =\mathrm{2}\right)\mathrm{P}\left(\mathrm{X}_{\mathrm{2}} =\mathrm{2}\right) \\ $$$$+\mathrm{P}\left(\mathrm{X}_{\mathrm{1}} =\mathrm{3}\right)\mathrm{P}\left(\mathrm{X}_{\mathrm{2}} =\mathrm{3}\right) \\ $$$$=\:\frac{\mathrm{C}_{\mathrm{0}} ^{\mathrm{3}} \mathrm{2}^{\mathrm{0}} \mathrm{C}_{\mathrm{0}} ^{\mathrm{4}} +\mathrm{C}_{\mathrm{1}} ^{\mathrm{3}} \mathrm{2}^{\mathrm{1}} \mathrm{C}_{\mathrm{1}} ^{\mathrm{4}} +\mathrm{C}_{\mathrm{2}} ^{\mathrm{3}} \mathrm{2}^{\mathrm{2}} \mathrm{C}_{\mathrm{2}} ^{\mathrm{4}} +\mathrm{C}_{\mathrm{3}} ^{\mathrm{3}} \mathrm{2}^{\mathrm{3}} \mathrm{C}_{\mathrm{3}} ^{\mathrm{4}} }{\mathrm{27}.\mathrm{16}} \\ $$$$=\:\frac{\mathrm{1}+\mathrm{24}+\mathrm{72}+\mathrm{32}}{\mathrm{432}}\:=\:\frac{\mathrm{43}}{\mathrm{144}}\:\approx\:\mathrm{29},\mathrm{9\%}. \\ $$

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