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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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
[tex]p = \frac{1}{2} = 0.5[/tex]
You are going to play the game twice.
This means that [tex]n = 2[/tex]
The probability that you win at least $1 both times is
This is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,2}.(0.5)^{2}.(0.5)^{2} = 0.25[/tex]
The probability that you win at least $1 both times is 0.25 = 25%.
The probability that you win at least $ 1 both times is 25%.
Since in a particular game, a fair die is tossed, and if the number of spots showing is either four or five, you win $ 1, while 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, and you are going to play the game twice, to determine the probability that you win at least $ 1 both times the following calculation must be performed:
Therefore, the probability that you win at least $ 1 both times is 25%.
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