kylebutler said:Homework Statement
z = 2, P(Z=z)=1/6
z = 3, P(Z=z)=1/6
z = 5, P(Z=z)=1/6
z = 7, P(Z=z)= x
z = 11, P(Z=z)= y
I'm supposed to find x and y given that E(Z)= 5+2/3
I have no idea how to do this. All I got is 7x+11x=4 but I can't solve this
LCKurtz said:I think you mean you got 7x+11y = 4. To get another equation remember that the Z probabilities must add to 1.
The expected value in probability is a measure of the average outcome of a random event, taking into account the probabilities of all possible outcomes.
To calculate the expected value in probability, you multiply each outcome by its probability and then add all of these products together. The formula is: E(x) = Σ x * P(x), where x represents each possible outcome and P(x) represents the probability of that outcome.
The expected value tells us what we can expect to happen on average if we repeat a random experiment many times. It is a useful tool for making predictions and decisions based on probability.
The expected value is not necessarily equal to the actual outcome of a single experiment. It is a long-term average based on probability. However, as the number of experiments increases, the actual outcome should approach the expected value.
Yes, the expected value can be negative. This means that on average, the outcome of a random experiment will result in a net loss. It is important to consider the expected value when making decisions involving risk and uncertainty.