Total number of outcomes = 36 The possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6). Let p be the probability of success, therefore, p = $\frac{6}{36}$ = $\frac{1}{6}$ So, q = 1 - p = 1 - $\frac{1}{6}$ = $\frac{5}{6}$ Since the dice is thrown 4 times, n=4 Let X denote the number of times of getting doublets in the experiment of throwing two dice simultaneously four times. Therefore X can take the values 0, 1, 2, 3, or 4. P (X = 0) = $\,{}^{4}C_0 p^0 q^4$ = $\left(\frac{5}{6}\right)^4$ = $\frac{625}{1296}$ P(X = 1) = $\,{}^{4}C_1 p^1 q^3$ = 4 $\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^3$ = $\frac{500}{1296}$ P(X = 2) = $\,{}^{4}C_2 p^2 q^2$ = 6 $\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2$ = $\frac{150}{1296}$ P (X = 3) = ${}^{4}C_3 p^4 q^0$ = $\left(\frac{1}{6}\right)^4$ = $\frac{1}{1296}$ Thus, the probability distribution is: