Find the mean and variance of the random variable X which denotes the number of doublets in four throws of a pair of dice.

 

 

 

 
 
 
 
 

Answers (1)

 

P(Doublets)=\frac{5}{36}=\frac{1}{6}\: \: \: \: \: P(No \: Doublets)= P'(D)\\

                                                                                            =\frac{30}{36}=\frac{5}{6}

            

P\left ( x \right )= \frac{625}{1296}\: \: \frac{500}{1296}\: \: \frac{150}{1296}\: \: \frac{120}{1296}\: \: \frac{1}{1296}

 

Mean = \sum xP(x)= 0.\frac{625}{1296}+1.\frac{500}{1296}+2.\frac{150}{1296}+3.\frac{120}{1296}+4.\frac{1}{1296}

Mean=\frac{500+300+360+4}{1296}=\frac{864}{1296}=\frac{2}{3}

Variance=\sum x^2P(x)-\sum xP(x)

\sum x^2P(x)= \left (0 \right )^2.\frac{625}{1296}+1^2.\frac{500}{1296}+2^2.\frac{150}{1296}+3^2.\frac{120}{1296}+4^2.\frac{1}{1296}

\sum x^2P(x)= \frac{1296}{1296}=1

Variance=1-\left (\frac{2}{3} \right )^2= \frac{5}{9}

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