 usagirl007A

2021-09-25

Whether it is appropriate to use a binomial probability distribution to obtain the probability that a contestant wins the game exactly twice out of five times or not.
Given: Consider the provided information in which there are 6 identical boxes, out of which one contains a prize. A contestant who wins the game exactly twice out of five times. Alix Ortiz

Calculation: According to the provided information, the total number of trials (n) is 5, which are independent. There are two possible outcomes (winning or not winning). Hence, the probability of success, that is, of winning the game each time, is:
$p=\frac{1}{6}$
And, the probability of failure, that is, of not winning the game, is:
$q=1-p$
$=1-\frac{1}{6}$
$=\frac{5}{6}$
The number of trials (n) is 5, out of which the player wins the game twice, that is, $r=2$. The general formula of a binomial distribution for calculating probability is as follows:
$P\left(r\right)={C}_{n,r}{p}^{r}{q}^{n-r}$
Here, n is the number of trials, r is a random variable that represents the number of successes out of n trials, and p is the probability of success. Now, the probability that a contestant wins exactly twice out of five times can be calculated as:
$P\left(2\right)={C}_{\left(5,2\right)}{\left(\frac{1}{6}\right)}^{2}{\left(\frac{5}{6}\right)}^{3}$
$=10×{\left(0.166\right)}^{2}{\left(0.833\right)}^{3}$
$=0.1592$
Interpretation: The probability that a contestant wins the game twice out of 5 times is 0.1592. According to the provided information, it is appropriate to use binomial probability distribution.

Do you have a similar question?