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 of success on a single trial, n=9, x=2, p=0.35.
To find the probability of x successes given the probability of success on a single trial with a binomial distribution, we can use the binomial probability formula:
P(x) = (n choose x) * p^x * (1 - p)^(n - x)
Where:
P(x) is the probability of x successes
n is the number of trials
p is the probability of success on a single trial
x is the number of successes
In this case, n = 9, x = 2, and p = 0.35.
Let's plug in these values and calculate the probability:
P(2) = (9 choose 2) * 0.35^2 * (1 - 0.35)^(9 - 2)
Calculating (9 choose 2):
(9 choose 2) = 9! / (2! * (9 - 2)!)
= 9! / (2! * 7!)
Since 9! = 9 * 8 * 7!:
(9 choose 2) = 9 * 8 / 2
= 36
Now let's substitute these values back into the formula:
P(2) = 36 * 0.35^2 * (1 - 0.35)^(9 - 2)
Calculating 0.35^2:
0.35^2 = 0.1225
Calculating (1 - 0.35)^(9 - 2):
(1 - 0.35)^(9 - 2) = 0.65^7
≈ 0.1124
Now let's substitute the calculated values back into the formula:
P(2) = 36 * 0.1225 * 0.1124
Calculating 36 * 0.1225 * 0.1124:
P(2) ≈ 0.1508
Therefore, the probability of having exactly 2 successes in 9 trials, with a probability of success on a single trial being 0.35, is approximately 0.1508.
To find the probability of x successes using the binomial probability formula, we use the formula:
P(x) = (nCx) * p^x * (1-p)^(n-x)
Where:
- P(x) is the probability of getting x successes.
- nCx is the number of ways to choose x successes out of n trials, given by nCx = n! / (x! * (n-x)!)
- p is the probability of success on a single trial.
- (1-p) is the probability of failure on a single trial.
- ^ denotes exponentiation.
In this case, n = 9, x = 2, and p = 0.35. Plugging these values into the formula, we get:
P(2) = (9C2) * 0.35^2 * (1-0.35)^(9-2)
Now, we need to calculate the values of (9C2), 0.35^2, and (1-0.35)^(9-2) separately.
1. Calculating (9C2):
Using the formula nCx = n! / (x! * (n-x)!), we have:
9C2 = 9! / (2! * (9-2)!)
9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
2! = 2 * 1
(9-2)! = 7 * 6 * 5 * 4 * 3 * 2 * 1
Substituting the values into the formula, we get:
9C2 = (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / ((2 * 1) * (7 * 6 * 5 * 4 * 3 * 2 * 1))
2. Calculating 0.35^2:
Simply square the value of p:
0.35^2 = 0.35 * 0.35
3. Calculating (1-0.35)^(9-2):
Subtract p from 1, and raise the result to the power of (n-x):
(1-0.35)^(9-2) = (1-0.35)^7
Now, we can substitute the calculated values into the formula:
P(2) = (9C2) * 0.35^2 * (1-0.35)^(9-2)
Solve each part of the equation individually, and then multiply the results:
P(2) = (9C2) * (0.35^2) * (1-0.35)^7