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.

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.

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

And p is the probability of X happening.

In this problem we have that:

Five tosses, so

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

(a) twice

This is

(b) at most once

(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

We want

So

the answer is 11.2 (i know i’m late)

i have no idea just need points sorry

step-by-step explanation: