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 success given the probability p of success on a single trial.
1. by formula- n=9, x=2, p=0.35
2. by PDF- n=15, x=13, p=1/3
C(9,2) = 9!/[2!(7!)]
= 9*8/2 = 9*4 = 36
so 36(.35)^2(1-.35)^7
= 36 * .1225 * .049
= .216
I do not know why by PDF means (By probability distribution function?)
yes thank you I believe that's what it stands for...
Well, I suppose I could look it up in a table of the distribution for n = 15, x =13, p =.3333... by symmetry
it is the same as for n = 15, x = 2, p = .33333....
but I do not have a good table
I have only up to n = 10
so have o do it the same old way
C(15,13) = 105
so
105 * (1/3)^13 * (2/3)^2
= 2.927*10^-5
c(15,13)
Now maybe they mean using the normal function which is the limit of binomial for large n
u = n p = 15*1/3 = 5
s^2 = 5(2/3) = 10/3 = 3.333...
s = 1.83
z = (x-u)/s = (13-5)/1.83 = 4.38
f(z) = (1/sqrt(2pi) e^-(4.38^2/2)
= 2.72 * 10^-5
gee, not too far off
For a binomial distribution with parameters
n = 5 , p = 0.3 . Find the probabilities of
getting :
(i) Atleast 3 successes.
(ii) Atmost 3 successes.
n = 5 , p = 0.3
To find the probability of x successes using the binomial probability formula, you can use the formula:
P(x) = C(n, x) * p^x * (1-p)^(n-x)
Where:
- P(x) is the probability of x successes
- C(n, x) is the binomial coefficient, which represents the number of ways to choose x successes out of n trials, and can be calculated as C(n, x) = 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
- x is the number of successes
- n is the total number of trials
Now let's solve the given examples:
1. By Formula:
For this example, n = 9, x = 2, and p = 0.35.
Using the formula, we have:
P(2) = C(9, 2) * (0.35)^2 * (1-0.35)^(9-2)
To calculate C(9, 2), we can use the combination formula:
C(9, 2) = 9! / (2! * (9-2)!) = 9! / (2! * 7!) = (9 * 8) / (2 * 1) = 36
Substituting the values:
P(2) = 36 * (0.35)^2 * (0.65)^7 ≈ 0.206
Therefore, the probability of 2 successes, given that the probability of success on a single trial is 0.35 and the procedure is repeated 9 times, is approximately 0.206.
2. By PDF:
For this example, n = 15, x = 13, and p = 1/3.
Using the probability mass function (PDF) of the binomial distribution, we can calculate the probability directly.
The PDF of a binomial distribution is given by:
P(x) = C(n, x) * p^x * (1-p)^(n-x)
Substituting the values:
P(13) = C(15, 13) * (1/3)^13 * (2/3)^(15-13)
Calculating C(15, 13):
C(15, 13) = 15! / (13! * (15-13)!) = 15! / (13! * 2!) = (15 * 14) / (2 * 1) = 105
Substituting the values:
P(13) = 105 * (1/3)^13 * (2/3)^2 ≈ 0.143
Therefore, the probability of 13 successes, given that the probability of success on a single trial is 1/3 and the procedure is repeated 15 times, is approximately 0.143.