Question

Suppose that a pair of balanced dice are rolled 120 times, and let X denote the number of rolls on which the sum of the two numbers is 12. Use the Poisson approximation to approximate \({\bf{Pr}}\left( {{\bf{X = 3}}} \right){\bf{.}}\)

Short answer

The required probability is 0.2202.

Step-by-step solution

Given information

A pair of dice is rolled 120 times. Therefore,

\(n = 120\)

X is a random variable that denotes the number of rolls needed to get a sum of 12 on the two dice.

Define the probability of success

Total number of outcomes in the sample space is 36\(\left( {{6^2}} \right)\).

The probability that the sum on two dice is 12 only when 6 is obtained on both dice.

\(\begin{aligned}{}p &= \Pr \left( {{\rm{Sum}}\,\; = 12} \right)\\ &= \frac{1}{{36}}\end{aligned}\)

Therefore, X is a random variable that follows binomial distribution, that is,

\(X \sim Bin\left( {n = 120,p = \frac{1}{{36}}} \right)\)

Define the Poisson approximation

The binomial distribution is approximated to the Poisson distribution if the samples are large(n) and probability p such that,

\(\begin{array}{c}np < 5\\n\left( {1 - p} \right) < 5\end{array}\)

In this case,

\(\begin{array}{c}np = 120 \times \frac{1}{{36}}\\ = 3.33\end{array}\)

The value is less than 5. Thus, the binomial distribution could be approximated to Poissonwith\(\lambda = 3.33\).

Thus,\(X \sim Poisson\left( {\lambda = 3.33} \right)\)

Calculate the required probability

Therefore, the probability is computed as follows,

\(\begin{aligned}{}\Pr \left( {X = 3} \right) &= \frac{{{e^{ - \lambda }} \times {\lambda ^x}}}{{x!}}\\ &= \frac{{{e^{ - 3.33}} \times {{3.33}^3}}}{{3!}}\\ &= 0.2202\end{aligned}\)

Hence the required probability is 0.2202