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  • Find the probability distribution of the random variable X, which denotes the number of doublets in four throws of a pair of dice. Hence, find the mean of the number of doublets (X).    </div

Find the probability distribution of the random variable X, which denotes the number of doublets in four throws of a pair of dice. Hence, find the mean of the number of doublets (X).

 

 

Answers (1)

 Let X: Number of doublets

X = 0,1,2,3,4

Total number of possible outcomes when we through a pair of dice = 36

Doublets = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) } 

 \\ \mathrm{P}(\text { Doublet })=\frac{6}{36}= \frac{1}{6} \\\\ \mathrm{P}(\text { Not a doublet })=\frac{5}{6}

X P(X) X.P(X)
0 \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} =\frac{625}{1296} 0
1 4 \times \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} =\frac{500}{1296} \frac{500}{1296}
2 6 \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} =\frac{150}{1296} \frac{300}{1296}
3 4 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} =\frac{20}{1296} \frac{60}{1296}
4 \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} =\frac{1}{1296} \frac{4}{1296}

\\ \text { Mean }=\sum X \cdot P(X) \\\\ =0+\frac{500}{1296} + \frac{300}{1296} + \frac{60}{1296}+\frac{4}{1296} \\\\ = \frac{864}{1296}\\\\ = \frac{2}{3}

Posted by

Safeer PP

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