The expected gain or loss exists estimated by (P(x) × n) then there exists a loss of $2.
When determining the mean of a probability distribution, we use expected value. The average value we anticipate to see prior to any data collection is represented by this. When attempting to determine the average value of a given sample, the mean is frequently utilized.
The probability of the 4 prizes which exists $500, $100, $25, and no prize.
P ($500 prize) = 1/100 or 0.01
P ($100 prize) = 2/100 or 0.02
P ($25 prize) = 4/100 or 0.04
P (No prize) = 100/100 – 1 + 2 + 4/100 = 93/100 = 0.93
Expected gain or loss exists estimated by: (P(x) × n)
substitute the values in the above equation, we get
E = (500-10) × 0.01 + (100-10) × 0.02 + (25-10) × 0.04 + (-10) × 0.93
simplifying the equation, we get
= 4.90 + 1.80 + 0.6 – 9.3
E = -2
There exists a loss of $2.
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