meaning multiplying an outcome by its probability I was reading about the expected value in probability theory. What is the meaning of multiplying an outcome by its probability. For example if I multiply and amount of money by an interest rate I get the iterest amount. What would mean to multiply an outcome by its probability?
 A: Say I am a basketball player that makes 50% of my shots, i.e. the probability of my shot going in is $P=0.5$.
Now suppose I try 100 shots.  How many shots can we expect to go in? 50, right?
Well, that is exactly $100*P$: $100*P = 100*0.5=50$
In terms of expected 'value': If all my shots that go in are worth 2 points, and my misses are worth 0 points, then I can expect to score how many points per shot?
That would be $0.5*2 + 0.5*0 = 1$
A: Suppose that there are only two outcomes of a random experiment: $1$ and $2$. Then consider the following outcomes of $10$ independent trials:
$$1,2,2,2,1,2,1,2,2,2.$$
You can calculate the relative frequencies of the $1$s and the $2$s
$$\frac3{10}\ \text{ and } \ \frac7{10},$$
respectively.
If you accept these values as probabilities (or at least very good approximations of probabilities) then, by definition, you would compute the  mean as
$$0.3\cdot1+0.7\cdot 2.$$
But if you do not know the concept of the mean then you would calculate the average as the commonsense counterpart of the mean:
$$\frac1{10}(1+2+2+2+1+2+1+2+2+2)=\frac1{10}(3\cdot 1+7\cdot 2)=\frac3{10}\cdot 1+\frac 7{10}\cdot 2$$
which is the same as the mean even if you don't know what the mean is. This is the intuitive basis of the definition of the mean.
A: If you're asking how an expected value works as a whole, see the other answers. An expected value is a sum of several terms, where each term is the product of an outcome with its probability.
I think you may be asking a very difficult question: What is the meaning of each individual term? Well, my answer is that each individual term is not necessarily meaningful!
Consider a train that is equally likely to arrive at 3:00 PM or 3:30 PM. What is its expected arrival time? Clearly 3:15 PM. But how did we calculate that? Let $x=$ 3:00 PM and $y=$ 3:30 PM be the two possible arrival times; these are inherently points in the affine space of all possible points in time. The expected value that we so easily computed was $\frac12 x + \frac12 y$. But what sort of object is $\frac12 x$, and what is its value?
In practice, we need a way to identify points with numbers. We could identify every point in time $t$ with the number of minutes that elapse between 3:00 PM and $t$. That would be very convenient for this example. Then we can write $x=0$ and $y=30$, so $\frac12 x + \frac12 y = \frac12\cdot0+\frac12\cdot30=0+15=15$. Okay, that was easy to compute. It may even be something like what you did in your head. But what does each term mean? Does $\frac12 x=0$ represent 3:00 PM, while $\frac12 y=15$ represents 3:15 PM? Surely not: did we really add together 3:00 PM and 3:15 PM to get the final answer?
Or we could identify every point in time $t$ with the number of minutes that elapse between noon and $t$. That's a bit more natural. You can try it, and you'll get 3:15 PM again as the final answer. But what about the terms? We now have $x=180$ and $\frac12 x=90$. Can we say that $\frac12 x$ represents 1:30 PM?
Or perhaps we should switch to a 24-hour clock and measure minutes from midnight. Then $x=900$ and $\frac12 x=450$. Is that the same as 7:30 AM??
Or perhaps we should switch to Unix time and identify a point in time with the seconds elapsed since 1970. Now we can't even make progress unless we know today's date...!
The answer is that none of the above are really correct, and $\frac12 x$ is not a time at all. It's secretly a member of some higher-level mathematical abstraction that sits on top of the affine space. Generally, a linear combination of points in an abstract affine space is not a member of that space unless the weights sum up to $1$, making it an affine combination. When the weights sum up to $1$ -- when you include every probability in your expected value computation -- then and only then, you get a result having the same meaning as the outcomes you started with, and the zero point doesn't matter.
Very often, you'll be dealing with spaces that do have a natural zero point available. When you talk about amounts of money, there's certainly a well-understood meaning of \$0, which is the same in any currency. Just be careful: the choice isn't always so obvious. For example, say you're computing an expected return on investment, and you're trying to determine how much each term really contributes to the expected value. Do you set the neutral ROI at $0$, or $1$, or the rate of inflation? Whatever you choose, be aware that your choice makes a difference.
A: Lets say we were betting on a dice roll. If you roll a $6$, you get $\$6$, but if you roll anything else you lose $\$3$. Should you take the bet? We can find out if we look at the expected value $$Value=(6*P(x=6))+(-2*P(x\not=6))$$ The first term is $1$, since the probability of rolling a $6$ is $\frac16$. The second term is $-2.5$, since the probability of not rolling a $6$ is $\frac56$. This gives an expected value of $-\$1.5$, meaning if you took the bet, you should expect to lose, on average, $\$1.5$. When I say on average, I mean if we played this game a large number of times, say $n$ times, you will have lost $1.5*n$ dollars.
A: To multiply an outcome by its probability means to take the value of potential outcome(s) and multiply those value(s) by the decimal analogous to the percent chance of the outcome(s) occurring in order to determine how much value you are likely to get.
Here are three examples that might help
Example 1:
You have ten dollars and a 100% chance of the money increasing by 50% in one year.
Then your expected value after 1 year is 15*1.00=15. 
Example 2:
Suppose you have $100  with a 60% chance of that money staying the same after 1 year and 20% chance of that money doubling in one year. The expected value is then 100*.6 + (2*100)*.2=$100 expected value after one year.
Example 3:
Let's say you have three bags. Bag A,B and C.
Each bag contains marbles some of which are black. You are told that if you grab a handful of marbles from each bag. Bag A has a 30% chance of giving you 3 black marbles in 1 handful, Bag B has a 50% chance of giving you 4 black marbles in 1 handful and bag C has 20% chance of giving you 5 marbles in one handful.
Now suppose as an experiment you grab exactly 1 handful of marbles from each of the three bags. The expected value of black marbles you will have after the experiment is 3.9. This was calculated as (.30 * 3) + (.50 * 4)  + (.20 *5)=3.9
