Question

In a particular game, a fair die is tossed. If the number of spots showing is either four or five, you win $1. If the number of spots showing is six, you win $4. And if the number of spots showing is one, two, or three, you win nothing. You are going to play the game twice. The probability that you win at least $1 both times is

Answer 1

Answer:

The probability that you win at least $1 both times is 0.25 = 25%.

Step-by-step explanation:

For each game, there are only two possible outcomes. Either you win at least $1, or you do not. Games are independent. This means that the binomial probability distribution is used to solve this question.

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.

Probability of winning at least $1 on a single game:

The die has 6 sides.

If it lands on 4, 5 or 6(either of the three sides), you win at least $1. So

$p = \frac{1}{2} = 0.5$

You are going to play the game twice.

This means that $n = 2$

The probability that you win at least $1 both times is

This is P(X = 2).

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

$P(X = 2) = C_{2,2}.(0.5)^{2}.(0.5)^{2} = 0.25$

The probability that you win at least $1 both times is 0.25 = 25%.