PLEASE HELP ! Someone has offered you a wager on the outcome of a die roll. You will pay him
$100 for each time the standard, six-sided die lands on one, and he will pay you $30 for each time the die lands on any other number. Without your awareness, he has loaded the die so that it has a 40% change to land on the number one, and besides
that an equal change to land on any of the values 2 through 6.

What is the expected value of one round of gambling in this scenario?

-$12
-$32
$2
-$22

You'll lose on average $22 per roll.

Explanation:

Normally there is a 1/6 chance to land on any given side of a standard die, but your friend has loaded the die in a way to make it have a 40% chance to land on "1" and an equal chance to land on anything else. Since there's a 40% chance to land on "1", this leaves 100% - 40% = 60% for everything else.  

Let's define two events

So far we know that P(A) = 0.40 and P(B) = 0.60; I'm using the decimal form of each percentage.

The net value of event A, which I'll denote as V(A), is -100 since you pay $100 when event A occurs. So we'll write V(A) = -100. Also, we know that V(B) = 30 and this value is positive because you receive $30 if event B occurs.

To recap things so far, we have the following:

Multiply the corresponding probability and net value items together

Then add up those products:

-40+18 = -22

This is the expected value, and it represents the average amount of money you earn for each dice roll. So you'll lose on average about $22. Because the expected value is not zero, this means this game is not mathematically fair.

This does not mean that any single die roll you would lose $22; instead it means that if you played the game say 1000 or 10,000 times, then averaging out the wins and losses will get you close to a loss of $22.

find out the value of b

step-by-step explanation:

i think asking google will you get the answer to that question