Probability & Statistics Ch.2 Section 3 Question 14

· Mohammad-Ali Bandzar

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Suppose that a symmetrical die is rolled 20 independent times, and each time we record whether or not the event {2,3,5,6}\{2, 3, 5, 6\} has occurred.

a) What is the distribution of the number of times this event occurs in 20 rolls? b) Calculate the probability that the event occurs five times.

Solution

(a) What is the distribution of the number of times this event occurs in 20 rolls?

This is asking for the binomial probability as all our trials are independent.

From Example 2.3.3 in the textbook:

pY(y)=P(Y=y)=(ny)θy(1θ)nyp_Y(y) = P(Y=y) = \binom{n}{y}\theta^y(1-\theta)^{n-y}

Where:

  • n = number of trials
  • θ = probability of success
  • y = desired number of successful outcomes

A die has 6 possible outcomes, all equally likely. We have 4 desired outcomes in this question, so:

θ=46=23\theta = \frac{4}{6} = \frac{2}{3}

We have 20 independent trials, so n = 20.

YBinomial(20,2/3)Y \sim \text{Binomial}(20, 2/3)

(b) Calculate the probability that the event occurs five times

P(Y=y)=(ny)θy(1θ)nyP(Y=y) = \binom{n}{y}\theta^y(1-\theta)^{n-y}

P(Y=5)=(205)(23)5(13)15P(Y=5) = \binom{20}{5}\left(\frac{2}{3}\right)^5\left(\frac{1}{3}\right)^{15}

=15504×32243×114348907= 15504 \times \frac{32}{243} \times \frac{1}{14348907}

P(Y=5)=49612834867844010.000142P(Y=5) = \frac{496128}{3486784401} \approx 0.000142