If X and Y are independent binomial variates `A(5,(1)/(2)) and B(7, (1)/(2))`. Find the value of P(X+Y=3).

Options

A. `35//47`
B. `55//1024`
C. `220//512`
D. `11//204`

Answer

B

Explanation

We, have
`P(X=r)=.^5C_(r)((1)/(2))^5 and , P(Y=r)=.^7C_r((1)/(2))^7`
Now,
`P(X+Y)=3)=P(X=0,Y=3)+P(X=1,Y=2)+P(X=2,Y=1)+P(X=3,Y=0)`
`=P(X=0)P(Y=3)+P(X=1)+P(Y=2)+P(X=2)P(Y=1)+P(X=3)P(Y=0)`
`=.^5C_0xx((1)/(2))^5xx.^7C_3xx((1)/(2))^7+.^5C_1xx((1)/(2))^5xx.^C_2xx((1)/(2))^7`
`+.^5C_(2)xx((1)/(2))^5xx.^7C_1xx((1)/(2))^7+.^5C_3xx((1)/(2))^5xx.^7C_0xx((1)/(2))^7`
`=((1)/(2^12))(35+5xx21+10xx7+10xx1)=(220)/(2^12)=(55)/(1024)`