Question: A box contains 10 tickets, 2 of which carry a prize of 7 8 each, 5 of which carry a prize of 3 4 each, and remaining 3 carry a prize of ? 2 each. If one ticket is drawn at random, find the mean value of the prize.
The correct answer is –
To find the mean value of the prize, we need to first find the probability of drawing each type of ticket and then calculate the expected value of the prize.
The probability of drawing a ticket with a prize of 7 8 is:
P(7 8) = 2/10 = 1/5
The probability of drawing a ticket with a prize of 3 4 is:
P(3 4) = 5/10 = 1/2
The probability of drawing a ticket with a prize of ? 2 is:
P(? 2) = 3/10
Now we can calculate the expected value of the prize:
E(prize) = P(7 8) x 7 8 + P(3 4) x 3 4 + P(? 2) x ? 2 E(prize) = (1/5) x 7 8 + (1/2) x 3 4 + (3/10) x ? 2 E(prize) = 7/20 + 3/8 – 3/50 E(prize) = 0.35 + 0.375 – 0.06 E(prize) = 0.665
Therefore, the mean value of the prize is 0.665.
