A pair of dice is thrown 4 times If getting a doublet is considered a success, find the probability - Maths - Probability

Question

A pair of dice is thrown 4 times If getting a doublet is
considered a success, find the probability of the number of
successes and hence find its mean

Ajanta Trivedi · Accepted Answer

throwing a doublet i e {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)} total number of outcomes = 36 let p be the probability of success therefore p=6/36=1/6 let q be the probability of failure therefore q= 1-p=1-1/6=5/6 since the dice is thrown 4 times , n=4 let X be the random variable for getting doublet, therefore X can take at max 4 values P(X=0)= ${}^{4}{C_{0}} p^{0} q^{4} = (\frac{5}{6})^{4} = \frac{625}{1296}$ P(X=1)= ${}^{4}{C_{1}} p^{1} q^{3} = 4 * \frac{1}{6} * (\frac{5}{6})^{3} = \frac{500}{1296}$ P(X=2)= ${}^{4}{C_{2}} p^{2} q^{2} = \frac{4 * 3}{2} * (\frac{1}{6})^{2} * (\frac{5}{6})^{2} = \frac{6 * 25}{1296} = \frac{150}{1296}$ P(X=3)= ${}^{4}{C_{3}} p^{3} q^{1} = 4 * (\frac{1}{6})^{3} * (\frac{5}{6})^{1} = \frac{20}{1296}$ P(X=4)= ${}^{4}{C_{4}} p^{4} q^{0} = 1 * (\frac{1}{6})^{4} * (\frac{5}{6})^{0} = \frac{1}{1296}$ X 0 1 2 3 4 P(X) $\frac{625}{1296}$ $\frac{500}{1296}$ $\frac{150}{1296}$ $\frac{20}{1296}$ $\frac{1}{1296}$ mean $μ = \sum_{i = 1}^{4} X_{i} P (X_{i}) = 0 * \frac{625}{1296} + 1 * \frac{500}{1296} + 2 * \frac{150}{1296} + 3 * \frac{20}{1296} + 4 * \frac{1}{1296} = \frac{500 + 300 + 60 + 4}{1296} = \frac{864}{1296} = \frac{54}{81} = \frac{2}{3}$ thus the mean is 2/3 hope this helps you cheers!!

X	0	1	2	3	4
P(X)	$\frac{625}{1296}$	$\frac{500}{1296}$	$\frac{150}{1296}$	$\frac{20}{1296}$	$\frac{1}{1296}$

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