Respuesta :
Answer:
The require expected value of x is [tex]\frac{9}{2}[/tex]
Step-by-step explanation:
Consider the provided information.
Mike will play a game in which he will toss a penny, a nickel, and a dime at the same time.
The probability of head is [tex]\frac{1}{2}[/tex]
He will be awarded 3 points for each coin that lands with heads faceup.
Let the random variable x represent the total number of points awarded on any toss of the coins.
Thus the expected value is:
[tex]\text{Expected value}=\sum x_i p(x_i)[/tex]
[tex]\text{Expected value}=3\times \frac{1}{2}+3\times \frac{1}{2}+3\times \frac{1}{2}[/tex]
[tex]\text{Expected value}=\frac{9}{2}[/tex]
Hence, the require expected value of x is [tex]\frac{9}{2}[/tex]
The expected value of X representing the average of the total number of points awarded on toss of the coins is 27/8
How to find the mean (expectation) of a random variable?
Supposing that the considered random variable is discrete, we get:
[tex]Mean = E(X) = \sum_{\forall x_i} f(x_i)x_i\\\\[/tex]
where [tex]x_i; \: \: i = 1,2, ... ,n[/tex] is its n data values
and [tex]f(x_i)[/tex] is the probability of [tex]X = x_i[/tex]
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}} = \dfrac{n(E)}{n(S)}[/tex]
where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
For this case, we're given that:
- 3 coins tossed, and its 3 points on each head outcome
- X = total number of points awarded
Now, X can have total 4 values, viz 0, 1, 2, and 3
Their each's probability is evaluated as:
- Case 1: No heads
That means all 3 coins needs to land on tail.
There is, therefore, single way to do so. Whereas, total outcomes possible are[tex]2\times 2\times 2 = 8[/tex] (by product rule, as there are 2 outcomes possible for each coin)
Thus, we get:
[tex]P(X = 0) = \dfrac{1}{8}[/tex]
- Case 2: Single head
When single head, that means 3 points, or X = 3
Total outcomes possible = 8,
Single head can occur in 3 different ways, as other two needs to be strictly tails.
Thus, [tex]P(X = 3) = \dfrac{2}{8}[/tex]
- Case 3: Two heads
Two heads means 3+3 = 6 points, or X = 6
Total outcomes possible = 8
Two heads means single tail. Single tail can occur in 3 different ways.
Therefore, [tex]P(X = 6) = \dfrac{2}{8}[/tex]
- Case 4: Three heads
Three heads means 3+3+3 = 9 points, or X = 9
Total outcomes possible = 8
Three heads means no tail, only single way to do it.
Thus, [tex]P(X = 9) = \dfrac{1}{8}[/tex]
Therefore, the expectation of X is calculated as:
[tex]E(X) = 0.\dfrac{1}{8} + 3.\dfrac{2}{8} + 6.\dfrac{2}{8} + 9.\dfrac{1}{8} = \dfrac{27}{8}[/tex]
Thus, the expected value of X is 27/8.
Learn more about expectation here:
https://brainly.com/question/4515179
