Question 5.15: A fair die is rolled 7 times. Find the probability that the ......
A fair die is rolled 7 times. Find the probability that the number 1 appears twice and the number 2 appears once.
The problem asks for the numbers 1 and 2. All the other four numbers (i.e., 3 through 6) are lumped into one event. Thus, instead of the normal 6 events, we have three events. Let K_1 be the number of times the number 1 appears in the 7 trials, K_2 the number times the number 2 appears in the 7 trials, and K_3 the number of times that all the other four numbers (3, 4, 5, 6) appear in the 7 trials. Similarly, let p_1 denote the probability that 1 appears in any roll of the die, p_2 the probability that 2 appears in any roll of the die and p_3 the probability that any number other than 1 and 2 appears in any roll of the die. Since the die is fair, we have that p_1=p_2=1/6 and p_3=1-p_1-p_2=4/6. Thus, the desired result is
\,\displaystyle p_{K_1 K_2 K_3}\left (2,1,4\right )=\binom{7}{214} p_1^2 p_2^1 p_3^4=\frac{7 !}{2 ! 1 ! 4 !}\left (\frac{1}{6}\right )^2\left (\frac{1}{6}\right )\left (\frac{4}{6}\right )^4=\frac{\left (105\right )\left (256\right )}{6^7}\\ \,\displaystyle =\frac{26,880}{279,936}=0.0960