Question

Imagine yourself at a fair playing one of the midway games. Pick a game and calculate the expected value and post your results along with how you calculated them. For example, you may decide to throw a basketball to try to win a $10 bear. You paid $2.00 for three shots. What is the expected value?

Answer 1

Answer:

The expected value of x= E(x) = np= 3 *1/2= 3/2= 1.5

Variance = npq= 3*1/2*1/2=3/4= 0.75

Step-by-step explanation:

In the given question the probability to win is equal to probability to loose. therefore P+q= 1 or p=q or p=1/2 and q= 1/2

The number of shots or trials are three. Therefore n= 3

Let x be the random variable with the binomial distribution b(x;n,p) . Then its mean and variance are calculated as follows

Here p =1/2 and q =1/2

n= 3

The expected value of x= E(x) = np= 3 *1/2= 3/2= 1.5

Variance = npq= 3*1/2*1/2=3/4= 0.75

Answer 2

The expected value of x= E(x) = np= $3 \times 1\div 2= 3\div 2= 1.5$

Calculation of expected value:

Since

The probability to win is equivalent to the probability to lose.

So,  P+q= 1 or p=q or p=1/2 and q= 1/2

Also, The number of shots or trials is three. So,  n= 3

Let us assume x be the random variable with the binomial distribution b(x;n,p) .

Now

Here p =1/2 and q =1/2

n= 3

So,

The expected value of x= E(x) = np= $3 \times 1\div 2= 3\div 2= 1.5$

Learn more about value here: https://brainly.com/question/22375402