Question

if you roll a pair of dice 100 times over the course of a game, what is the probability that you will have rolled doubles, at any point, exactly three times?
a. 0.167
b. 0.5
c. 0.00463
d. 0.0000156
please select the best answer from the choices provided
a
b
c
d

Explanation:

Step1: Calculate the probability of rolling doubles in a single roll

When rolling a pair of dice, there are \(n(S)=6\times6 = 36\) possible outcomes. The number of doubles \((1 - 1,2 - 2,\cdots,6 - 6)\) is \(n = 6\). So the probability of rolling doubles in a single roll \(p=\frac{6}{36}=\frac{1}{6}\), and the probability of not rolling doubles \(q = 1 - p=1-\frac{1}{6}=\frac{5}{6}\)

Step2: Use the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times q^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successful trials, \(p\) is the probability of success in a single trial, \(q\) is the probability of failure in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Here, \(n = 100\), \(k = 3\), \(p=\frac{1}{6}\), \(q=\frac{5}{6}\)

First, calculate \(C(100,3)=\frac{100!}{3!(100 - 3)!}=\frac{100\times99\times98}{3\times2\times1}=161700\)

Then, \(p^{3}=(\frac{1}{6})^{3}=\frac{1}{216}\), \(q^{n - k}=(\frac{5}{6})^{97}\)

\(P(X = 3)=C(100,3)\times(\frac{1}{6})^{3}\times(\frac{5}{6})^{97}\)

\(\approx161700\times\frac{1}{216}\times(0.00006)\)

\(P(X = 3)\approx0.00463\)

Answer:

C. 0.00463