Question

Stacy uses a spinner with six equal sections numbered 2, 2, 3, 4, 5, and 6 to play a game. Write a probability model for this experiment, and then use the model to predict how many times Stacy would spin a 6 if she spun 50 times. Give the probabilities as decimals, rounded to 2 decimal places.

Answer 1

Hence, Stacy will spin 6, 8.33 times out of her n = 50 attempts.

Explanation:

Solution:-

- First thing we need to note that the probability of getting any number 1 , 2 , 3 , 4 , 5 or 6 is independent from each successive n trials.

- We will assume spinner to be unbiased, i.e the section of each number has the same dimensions (area). Then the probability to get any number of the spinner is = 1 / 6.

- We will define a random variable X : The number of times Stacy gets a 6 on n = 50 trials of spin.

- The random variable X has the probability of success p = the probability to get a 6 on each trial is 1/6. The probability remains same for each trial, so we can assume that RV X follows binomial distribution:

X ~ Bin ( n , p )

X ~ Bin ( 50 , 1/6 )

- The pmf of the Binomial distribution is given by:

P ( X = r ) = n C r * ( p )^r * ( 1 - p )^( n - r )

Where, r = The number times she gets a 6 on her spins.

- The expected number of times she Stacy gets a 6 for n = 50 trials is given:

E ( X ) = n*p

= 50 * 1/6

= 8.33 times

Hence, Stacy will spin 6, 8.33 times out of her n = 50 attempts.

Answer 2

We estimate to have 8.33 times the number 6 in 50 trials.

Explanation:

Let us consider a success to get a 6. In this case, note that the probability of having a 6 in one spin is 1/6. We can consider the number of 6's in 50 spins to be a binomial random variable. Then, let X to be the number of trials we get a 6 out of 50 trials. Then, we have the following model.

`P(X=k) = \binom{50}{k}((1)/(6))^k((5)/(6))^(50-k)`

We will estimate the number of times that she spins a 6 as the expected value of this random variable.

Recall that if we have X as a binomial random variable of n trials with a probability of success of p, then it's expected value is np.

Then , in this case, with n=50 and p=1/6 we expect to have
`(50)/(6)` number of times of having a 6, which is 8.33.