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

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i

9

step-by-step explanation: