A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of number of successes. Also, find the mean and variance of number of successes.

Answer

`(##RSA_MATH_XII_C31_E01_006_A01##)`
`Mean =(2)/(3), variance =(5)/(9)`

Explanation

In a single throw, P(doublet) `=(6)/(36)=(1)/(6),` and P(non-doublet)`=(1-(1)/(6)=(5)/(6).`
Let X be the number of doublets. Then, X = 0, 1, 2 or 3.
P(X=0)= P (non-doublet in each case)
`=P(barD_(1) barD_(2)barD_(3) barD_(4))=((5)/(6)xx(5)/(6)xx(5)/(6)xx(5)/(6))=(625)/(1296).`
P(X=1)=P(one doublet)
`=P(D_(1) barD_(2)barD_(3) barD_(4)) or P(barD_(1) D_(2)barD_(3) barD_(4))`
`or P(barD_(1) barD_(2)D_(3) barD_(4))orP(barD_(1) D_(2)barD_(3) D_(4))`
`=((1)/(6)xx(5)/(6)xx(5)/(6)xx(5)/(6))+((5)/(6)xx(1)/(6)xx(5)/(6)xx(5)/(6))+((5)/(6)xx(5)/(6)xx(1)/(6)xx(5)/(6))+((5)/(6)xx(5)/(6)xx(5)/(6)xx(1)/(6))`
`=(4xx(125)/(1296))=(125)/(324).`
P(X=2)=P(two doublets)
`=P(D_(1)D_(2) barD_(3)barD_(4))or P(D_(1)barD_(2)D_(3) barD_(4))or P(D_(1) barD_(2) barD_(3) D_(4)) or P(barD_(1) D_(2) D_(3) barD_(4)) or P(barD_(1) D_(2) barD_(3) D_(4)) or P(barD_(1) barD_(2) D_(3) D_(4))`
`=((1)/(6)xx(1)/(6)xx(5)/(6)xx(5)/(6))+((1)/(6)xx(5)/(6)xx(1)/(6)xx(5)/(6))+((1)/(6)xx(5)/(6)xx(5)/(6)xx(1)/(6))+((5)/(6)xx(1)/(6)xx(1)/(6)xx(5)/(6))+((5)/(6)xx(1)/(6)xx(5)/(6)xx(1)/(6))+((5)/(6)xx(5)/(6)xx(1)/(6)xx(1)/(6))`
`(6xx(25)/(1296))=(25)/(216).`
P(X=3)=P(three doublets)
`P(D_(1) D_(2) D_(3) barD_(4) or P(D_(1) D_(2) barD_(3) D_(4)) or P(D_(1) barD_(2) D_(3)D_(4)) or P(barD_(1) D_(2) D_(3) D_(4))`
`=((1)/(6)xx(1)/(6)xx(1)/(6)xx(5)/(5))+((1)/(6)xx(1)/(6)xx(5)/(6)xx(1)/(6))+((1)/(6)xx(5)/(6)xx(1)/(6)xx(1)/(6))+((5)/(6)xx(1)/(6)xx(1)/(6)xx(1)/(6))`
`=(4xx(5)/(1296))=(5)/(324).`
`P(X=4)=P("four doublets")=P(D_(1) D_(2) D_(3) D_(4))`
`=((1)/(6)xx(1)/(6)xx(1)/(6)xx(1)/(6))=(1)/(1296).`
Thus, we have

Now, find `mu and sigma^(2).`