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Question

# A die is tossed twice. A success is getting an odd number on a toss. Find the mean and the variance of the probability distribution of the number of successes.

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Solution

## Let x be the random variable denoting the number of times an odd number (the number of successes) when a die is tossed twice.Then x takes the values 0,1,2Let P(X=0) be probability of getting no odd number (both times showing even).∴P(X=0)=36×36=14Let P(X=1) be probability of getting odd number once.∴P(X=1)=2C136×36=66×36=12Let P(X=2) be probability of getting odd number twice.∴P(X=2)=36×36=14Thus the probability distribution of X is given by X=x x=0 x=1 x=2 P(X=x) 14 12 14We know that mean E(X)=∑xiPi=0×14+1×12+2×14∴E(X)=0+12+12=1Thus mean E(X)=1.We know that var(X)=E(X2)−[E(X)]2E(X2)=∑x2iPi=0×14+12×12+22×14∴E(X2)=0+12+4×14=32Thus var(X)=32−[1]2=32−1=12Hence mean is 1 and variance is 12.

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