A 6-sided die is loaded in such a way where odd outcomes occur 2/9 of the time, and even outcomes happen the other 1/9 of the time. What is this die’s expected value?

Respuesta :

Exact answer as a fraction is 10/3

Approximate answer in decimal form is 3.333

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Explanation:

Make a table as shown below. We have X as the result of the die, so X takes on values between 1 and 6. The second column is the probability P(X) column. If X is odd, then P(X) = 2/9. Otherwise, P(X) = 1/9

Then make a third column that multiplies the X and P(X) values. For instance, in row 1 we have X*P(X) = 1*1/9 = 1/9. Row 2 will have X*P(X) = 2*2/9. And so on.

Once you compute all the X*P(X) values, add up the results

EV = Expected value

EV = sum of the X*P(X) values

EV = 2/9 + 2/9 + 6/9 + 4/9 + 10/9 + 6/9

EV = (2+2+6+4+10+6)/9

EV = 30/9

EV = 10/3

EV = 3.333 approximately

Ver imagen jimthompson5910

The expected value of a 6-sided dies for the given event will be around 3.33.

What is probability?

The probability of an event occurring is defined by probability.

Probability is also known as chance because, if you flip a coin, the likelihood that it will land on its head or tail is nothing more than the chance that either the head or the tail will occur.

In our daily time, there are several instances in the everyday world where we may need to draw conclusions about how everything will turn out.

We know that the expected value is the sum of X[P(X)].

Given that at odd outcomes probability is = 2/9

At even outcomes probability is = 1/9

The value of X = 1,2,3,4,5,6

EV = sum of the X[P(X)] values

EV = 1(2/9) + 2(1/9) + 3(2/9) + 4(1/9) + 5(2/9) + 6(1/9)

EV = (2+2+6+4+10+6)/9

EV = 30/9

EV = 10/3

EV = 3.333 hence the expected value will be around 3.33.

For more information about the probability,

brainly.com/question/11234923

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