For a lottery game, each ticket has a probability of 1 / 3.09 = 0.324 of winning a prize. Every friday you buy 1 lottery game.
a) What is the probability you will win on two of your next three games?
p = p win = .324
1-p = p lose = .676
binary distribution
P(x = k) = C(n,k) p^k (1-p)^(n-k)
C(2,3) = 3!/2!(3-2)! = 3
so
= 3(.324)^2 (.676)
= .213
To find the probability of winning on two out of three games, we need to use the concept of binomial probability.
The binomial probability formula is:
P(X = k) = nCr * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability of exactly k successes
n is the number of trials
r is the number of successes
p is the probability of success on a single trial
In this case, our trials (n) are the three games you will play, the number of successes (r) is 2, and the probability of winning a single game (p) is 0.324.
Now, we can substitute these values into the formula and calculate the probability:
P(X = 2) = 3Cr * 0.324^2 * (1 - 0.324)^(3 - 2)
To find the binomial coefficient (nCr), we use the formula:
nCr = n! / (r! * (n - r)!)
Plugging in the values:
3Cr = 3! / (2! * (3 - 2)!)
= 3! / (2! * 1!)
= 3
P(X = 2) = 3 * 0.324^2 * (1 - 0.324)^(3 - 2)
Calculating further:
P(X = 2) = 3 * 0.105576 * 0.676
P(X = 2) ≈ 0.2269
Therefore, the probability of winning on two out of your next three games is approximately 0.2269 or 22.69%.