Football teams `T_(1)` and `T_(2)` have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of `T_(1)` winning,drawing and losing a game against `T_(2)` are `(1)/(2),(1)/(6)` and `(1)/(3)`, respectively. Each teams gets 3 points for a win, 1 point of a drawn and 0 point for a loss in a games.
`P(X=Y)` is
A. `(11)/(36)`
B. `(1)/(3)`
C. `(13)/(36)`
D. `(1)/(2)`
`P(X=Y)` is
A. `(11)/(36)`
B. `(1)/(3)`
C. `(13)/(36)`
D. `(1)/(2)`
Correct Answer – C
`P[X=Y]=P(“draw”).P(“draw”)+P(T_(1)”win”)P(T_(2)”win”)+P(T_(2)”win”).P(T_(1)”win”)`
`=(1//6xx1//6)+(1//2xx1//3)+(1//3xx1//2)=13//36`