Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. n=19, x=15 p=.85, find P(15)
To find the probability of exactly x successes (P(x)) in a binomial distribution, we can use the binomial probability formula:
P(x) = (nCx) * p^x * (1 - p)^(n - x)
Where:
- n is the number of trials
- x is the number of successes
- p is the probability of success on a single trial
- (nCx) represents the combination or "n choose x" which is calculated as n! / (x!(n - x)!)
In this case, n = 19, x = 15, and p = 0.85. We need to find P(15).
Using the provided values, the calculation becomes:
P(15) = (19C15) * (0.85)^15 * (1 - 0.85)^(19 - 15)
Now, let's break down each component of the formula:
1. (19C15) = 19! / (15!(19 - 15)!) = (19 * 18 * 17 * 16) / (4 * 3 * 2 * 1) = 38,760 / 24 = 1,615
2. (0.85)^15 = 0.085^2 * 0.85^13 = 0.00075 * (0.85^13) ≈ 0.00075 * 0.036 = 0.000027
3. (1 - 0.85)^(19 - 15) = 0.15^4 = 0.00050625
By substituting these values back into the formula, we have:
P(15) = 1,615 * 0.000027 * 0.00050625 ≈ 0.0213
Therefore, the probability of having exactly 15 successes (x = 15) in a binomial distribution with n = 19 trials and p = 0.85 is approximately 0.0213.