At the Statsville County Fair, the probability of winning a prize in the basketball toss game is 0.1.
a) Show the probability distribution for the number of prizes won in 8 games (round to 6 decimal places).
b) If the game will be played 500 times during the fair, how many prizes should the game operators keep in stock?
a) To show the probability distribution for the number of prizes won in 8 games, we can use the binomial distribution formula. The binomial distribution is used when there are only two possible outcomes (success or failure) for each trial, and the probability of success is constant.
In this case, the probability of winning a prize in each game is given as 0.1, which is the probability of success. The number of games played, n, is 8.
The binomial distribution formula is:
P(x) = (nCx) * p^x * q^(n-x)
Where:
P(x) is the probability of getting exactly x successes
nCx is the number of ways to choose x successes from n trials (combination)
p is the probability of success in one trial
q is the probability of failure in one trial (q = 1 - p)
x is the number of successes
Let's calculate the probability distribution for the number of prizes won in 8 games:
P(0) = (8C0) * (0.1)^0 * (1 - 0.1)^(8-0)
P(1) = (8C1) * (0.1)^1 * (1 - 0.1)^(8-1)
P(2) = (8C2) * (0.1)^2 * (1 - 0.1)^(8-2)
P(3) = (8C3) * (0.1)^3 * (1 - 0.1)^(8-3)
P(4) = (8C4) * (0.1)^4 * (1 - 0.1)^(8-4)
P(5) = (8C5) * (0.1)^5 * (1 - 0.1)^(8-5)
P(6) = (8C6) * (0.1)^6 * (1 - 0.1)^(8-6)
P(7) = (8C7) * (0.1)^7 * (1 - 0.1)^(8-7)
P(8) = (8C8) * (0.1)^8 * (1 - 0.1)^(8-8)
Let's calculate each probability:
P(0) = (8C0) * (0.1)^0 * (1 - 0.1)^(8-0) = 1 * 1 * 0.729 = 0.729
P(1) = (8C1) * (0.1)^1 * (1 - 0.1)^(8-1) = 8 * 0.1 * 0.6561 = 0.524288
P(2) = (8C2) * (0.1)^2 * (1 - 0.1)^(8-2) = 28 * 0.01 * 0.59049 = 0.164354
P(3) = (8C3) * (0.1)^3 * (1 - 0.1)^(8-3) = 56 * 0.001 * 0.531441 = 0.0403536
P(4) = (8C4) * (0.1)^4 * (1 - 0.1)^(8-4) = 70 * 0.0001 * 0.4782969 = 0.005505024
P(5) = (8C5) * (0.1)^5 * (1 - 0.1)^(8-5) = 56 * 0.00001 * 0.43046721 = 0.000360150375
P(6) = (8C6) * (0.1)^6 * (1 - 0.1)^(8-6) = 28 * 0.000001 * 0.387420489 = 1.65381715e-05
P(7) = (8C7) * (0.1)^7 * (1 - 0.1)^(8-7) = 8 * 0.0000001 * 0.34867844 = 2.68435456e-06
P(8) = (8C8) * (0.1)^8 * (1 - 0.1)^(8-8) = 1 * 0.00000001 * 0.313810596 = 3.13810596e-07
Rounded to 6 decimal places, the probability distribution for the number of prizes won in 8 games is:
P(0) = 0.729
P(1) = 0.524288
P(2) = 0.164354
P(3) = 0.040354
P(4) = 0.005505
P(5) = 0.000360
P(6) = 0.000017
P(7) = 0.000003
P(8) = 0.000000
b) To determine how many prizes the game operators should keep in stock, we need to consider the expected number of prizes won in 500 games.
The expected value or mean of a binomial distribution is given by the formula:
μ = n * p
Where:
μ is the expected value
n is the number of trials
p is the probability of success in one trial
In this case, the number of games played, n, is 500 and the probability of winning a prize in each game, p, is 0.1.
μ = 500 * 0.1 = 50
Therefore, the game operators should keep approximately 50 prizes in stock for the 500 games.