Answer:
P(Y=r) = [tex]4Cr (5/9)^r (4/9)^{4-r},r=0,1,2,3,4[/tex]
Step-by-step explanation:
Given that a bag contains 1 red, 3 green, and 5 yellow balls. A sample of four balls is picked, with replacement
When balls are picked with replacement, each time the ball picked up does not depend on the previous outcomes.
Thus given y=2 will not affect the pdf of G
In G, Y no of yellow balls can take values as 0,1,2,3,4
Prob of drawing a yellow ball each time = 5/9
Y = no of yellow balls is binomial with n =4 and p = 5/9
P(Y=0) = [tex](1-5/9)^4 = (\frac{4}{9} )^4[/tex]
In general,
P(Y=r) = [tex]4Cr (5/9)^r (4/9)^{4-r},r=0,1,2,3,4[/tex]