How do I plug this into the Binomial Probability Formula?
prob(lost) = 5/1000 = 1/200
prob(not lost) = 199/200
prob( 2 of 22 are lost)
= C(22,2) (1/200)^2 (199/200)^20
= .005224..
To plug this into the Binomial Probability Formula, you need to know the values for n, p, and k, where:
- n is the total number of trials (in this case, the number of items),
- p is the probability of success (in this case, the probability of an item being lost), and
- k is the number of successful events (in this case, the number of lost items).
In your example, n = 22 (since there are 22 items), p = 1/200 (the probability of an item being lost), and k = 2 (the number of lost items you want to find the probability for).
The Binomial Probability Formula is:
P(k) = C(n, k) * p^k * (1-p)^(n-k)
Now, let's substitute the values into the formula:
P(2) = C(22, 2) * (1/200)^2 * (199/200)^(22-2)
To calculate C(22, 2) (the number of ways to choose 2 items out of 22), we use the binomial coefficient formula:
C(22, 2) = 22! / (2! * (22-2)!)
Simplifying further:
C(22, 2) = 22! / (2! * 20!)
= (22 * 21 * 20!) / (2! * 20!)
= (22 * 21) / 2
= 11 * 21
= 231
Now we can substitute this value into the formula:
P(2) = 231 * (1/200)^2 * (199/200)^(22-2)
Calculating the result using a calculator or software, we get:
P(2) ≈ 0.005224