A dice game pays a player $5 for rolling a3 ot a 5 with a single die. the player has to pay $2 for any other roll. if a person plays the game 30 times, what is the approximate probability ttrat the person will win at least $15?

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

[tex]X \sim B(n,p)[/tex]

The probability that out of n trials, there'd be x successes is given by

[tex]P(X =x) = \: ^nC_xp^x(1-p)^{n-x}[/tex]

The expected value and variance of X are:

[tex]E(X) = np\\ Var(X) = np(1-p)[/tex]

Suppose that:

X = Random variable tracking the number of times out of 30 rolls of die that the player gets 3 or 5
Y = 50 - X = Random variable tracking the number of times that player gets 1,2,4 or 6 in those 30 rolls.

Then, the total money that the player would get is:

Z = 5X - 2Y

(as for each of X wins, the player gets $5 , else for each of Y loses, the player lose $2).

We've to get the probabiilty that Z ≥ 15

That means, we need:

[tex]5X -2Y \geq 15\\5X -2(30-X) \geq 15\\\\5X + 2X \geq 75\\\\X \geq 75/7 \approx 10.7\\\\X \geq 11[/tex]

Therefore, we get:

[tex]P(Z \geq 15) = P(X \geq 11)[/tex]

Now, assuming that each roll of die is independent of other rolls in terms of its result, we have all those 30 experiments independent.

Also, we can take:

Success in an experiment = Rolling 3 or 5 in that experiment

Failure in an experiment = Not rolling 3 or 5, or say that rolling 1,2,4 or 6 in that experiment.

Assuming all faces are equally likely, we get:

[tex]p = P(\text{Success in an experiment of roll of a die}) = \dfrac{2}{6} = 1/3[/tex] (for this case when success in an experiment is rolling 3 or 5 in that experiment)

Each experiment of those 30 independent experiments can have either success or failure (only 2 possible outcomes), so each of them are bernoulli trials.

Now, as we've got:

X = Random variable tracking the number of times out of 30 rolls of die that the player gets 3 or 5,

so, it means, X = total count of successes in those 30 bernoulli trials.

And therefore, we've got:

[tex]X \sim B(n = 30, p = 1/3)[/tex]

Using the probability function of binomial distribution, we get:
[tex]P(X \geq 11) = 1- P(X < 10) = \sum_{i=0}^{10} \: ^{30}C_i (1/3)^i (1- 1/3)^{30-i}[/tex]

Using calculator, we get:

[tex]P(X \geq 11) = \: ^{30}C_i (1/3)^i (1- 1/3)^{30-i} \approx 0.585[/tex]

Thus, we get:

[tex]P(Z \geq 15) = P(X \geq 11) \approx 0.585[/tex]

Thus, the probability that the player in this considered case would win at least $15 is found being 0.585 approximately.

Learn more about binomial distribution here:

https://brainly.com/question/13609688

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