In each turn of a game you toss two coins. If 2 heads come up, you win 2 points and if 1 head comes up you win 1 point. If no heads come up, you lose 3 points. What is the expected value of the number of points for each turn? –0. 25 0 0. 25 0. 5.

Expected value is the weighed sum, weights being the probabilities. The expected value of the number of points for each turn is 0.25

How to calculate the probability of an event?

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}}[/tex]

Where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

How to calculate the expectation of a discrete random variable?

Expectation can be taken as a weighted mean, weights being the probability of occurrence of that specific observation.

Thus, if the random variable is X, and its probability mass function is given : [tex]f(x) = P(X = x)[/tex] , then we have:

[tex]E(X) = \sum_{i=1}^n( f(x_i) \times x_i)[/tex]

(n is number of values X takes)

For the given case, let a random variable X track the number of points for each turn.

The values that X can assume is

X = 2 if 2 heads

X = 1 if 1 head

X = -3 if 0 heads.

In toss of two coins, we get the probability of obtaining heads as:

There are 4 results in toss of two coins (T,H), (H,T), (T,T), (H,H)

Where T shows tail as output and H is head as output.

Case 1: E= event of getting 2 heads in toss of two coins[tex]P(E) = 1/4 = 0.25[/tex] (as 1 favorable case (H,H) )
Case 2: E= event of getting 1 heads in toss of two coins [tex]P(E) = 2/4 = 0.5[/tex] (as 2 favorable cases (T,H), (H,T)
Case 3: E= event of getting 0 heads in toss of two coins [tex]P(E) = 1/4 = 0.25[/tex] (as 1 favorable case (T,T) )

The expectation is calculated as:

[tex]E(X) = P(X=0).-3 + P(X=1).1 + P(X=2).2= -0.75 +0.5 + 0.5 \\E(X) = 0.25[/tex]

Thus, the expected value of the number of points for each turn is 0.25

Learn more about expectation here:

https://brainly.com/question/4515179