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Question
Find the mean and variance of the random variable X which denotes the number of doublets in four throws of a pair of dice.
Solution
(i) Probability of getting a doublet in single throw = `6/36=1/6`
Probability of not geting doublet = 1 -`1/6=5/6`
Let x denotes a random variable which is number of doublets obtained in four throws of a pair of dice with n = 4.
P (X = x ) = nCx px qn-x
P(X = 0 ) = `"^4C_0 p^0 q^4 = (5/6)^4 = 625/1296`
`P(X = 1) ="^4 C_1 (1/6)^1(5/6)^3 = 500/1296`
`P(X = 2) ="^4C_2(1/6)^2(5/6)^2 = 150/1296`
`P(X = 3) ="^4C_3 (1/6)^3(5/6)^2=20/1296`
`P(X = 4) = "^4C_4 (1/6)^1 (5/6)^0 = 1/1296`
`E(X) = (0xx625)/1296 + (1xx500)/1296 +(2 xx 150)/1296 + (3 xx 20)/1296 + (4xx1)/1296`
`= 500/1296 + 300/1296 + 60/1296 + 4/1296`
` =216/324 = 54/81 =6/9 =2/3`
(ii) Variance = E ( X2 ) - (E(x))2
`E(X^2 ) = 0^2 xx 625/1296 + 1^2 xx (500/1296) +2^2 (150/1296) +3^2 (20/1296)+4^2(4/1296)`
`= 0+500/1296 +600/1296 +180/1296 +64/1296`
` = 1344/1296 = 336/324 =84/81 =28/27`
Variance `= 28/27 - 4/9 = (28-12)/27 =16/27`
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