# Explain whether the binomial probability distribution can be used to

Explain whether the binomial probability distribution can be used to find the probability that a contestant who plays the game five times wins exactly twice.
Check for the binomial experiment requirements and specify the values of n,r, and p.

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Calculation:
From the given information, it can be observed that the five trials are independent with two outcomes (win or do not win) that are same.
They are repeated under same conditions with the same probability of success on each trial.
Thus, the binomial distribution can be used to find the specified probability.
Values:
Number of trials $$\displaystyle{n}={5}$$
Probability of winning the game (or success) $$\displaystyle{p}={\frac{{{1}}}{{{6}}}}$$.
Number of times the player wins the game $$\displaystyle{r}={2}$$.
The probability of failure (or probability of not winning) is given below:
$$\displaystyle{q}={1}-{p}$$
$$\displaystyle={1}-{\left({\frac{{{1}}}{{{6}}}}\right)}$$
$$\displaystyle={\frac{{{5}}}{{{6}}}}$$
Binomial probability:
The probability of r successes out of n trials is given below:
$$\displaystyle{P}{\left({r}\right)}={C}_{{{n},{r}}}{p}^{{{r}}}{q}^{{{n}-{r}}}$$
Here, n is the number of trials, r is the number of successes, p is the probability of success, and q is the probability of failure.
The probability that a contestant wins exactly twice out of five times is calculated as given below:
$$\displaystyle{P}{\left({2}\right)}={C}_{{{5.2}}}{\left({\frac{{{1}}}{{{6}}}}\right)}^{{{2}}}{\left({\frac{{{5}}}{{{6}}}}\right)}^{{{3}}}$$
$$\displaystyle={\frac{{{5}!}}{{{2}!{3}!}}}\times{\left({0.166}\right)}^{{{2}}}\times{\left({0.833}\right)}^{{{3}}}$$
$$\displaystyle={10}\times{\left({0.166}\right)}^{{{2}}}\times{\left({0.833}\right)}^{{{3}}}$$
$$\displaystyle={0.160751}$$
Therefore, the probability that a contestant wins exactly twice is 0.160751.