Question

Find the probability that in five tosses of a fair die, a 3 will appear (a) twice, (b) at most once, (c) at least two times.

Answer 1

Answer:

A) 0.1612

B) 0.8031

C) 0.1969

Step-by-step explanation:

For each toss of the die, there are only two possible outcomes. Either it is a 3, or it is not. The probability of getting a 3 on each toss is independent from other tosses. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

Five tosses, so $n = 5$

The die has 6 values, from 1 to 6. The die is fair, so each outcome is equally as likely. The probability of a 3 appearing in a single throw is $p = \frac{1}{6} = 0.167$

(a) twice

This is $P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{5,2}.(0.167)^{2}.(0.833)^{3} = 0.1612$

(b) at most once

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.167)^{0}.(0.833)^{5} = 0.4011$

$P(X = 1) = C_{5,1}.(0.167)^{1}.(0.833)^{1} = 0.4020$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.4011 + 0.4020 = 0.8031$

(c) at least two times.

Either a 3 appears at most once, or it does at least two times. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 1) + P(X \geq 2) = 1$

We want $P(X \geq 2)$

So

$P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.8031 = 0.1969$