binomial distributions

Question

Find the probability of 3 successes in 7 trials of a binomial experiment in which the probability of success in any one trial is 35%. Round to the nearest tenth of a percent.

Answer 1

The probability of X successes in Y trials in which the probability of success in any trial is Z% is calculated as based on the product of:

(1) the number of combinations of success,

(2) percentage of success^number of successes,

(3) percentage of failures^number of failures

Formula: YCx x (Z%)^X x (1-Z%)^(Y-X)

: 7C3 x (0.35)^3 x (0.65)^4

: 7!/3!4! x (0.35)^3 x (0.65)^4

: 35(0.35)^3(0.65)^4

Answer = 26.8%

Answer 2

Hello Grace,

The binomial probability distribution is given by

p(X) = n!/[x!(n-x)!] p^x (1 - p)^n-x ,

where n is the number of trials, x is the number of successes, and p is the probability of success on any given trial. Recall that in the above formula, ! represents the factorial function. (For example, 5! = 5*4*3*2*1 = 120.) In your question, n =7, X = 3, and p = .35. Therefore, the desired probability is

p(3) = 7!/[3!(7 - 3)!] .35³ (1 - .35)^7-3,

p(3) = 7!/[3!*4!] .35³ (.65)⁴

p(3) = 5040/[6*24] .35³ (.65)⁴

p(3) = .26787

Expressed in percent terms, the probability is 26.8% (rounded to the nearest tenth of a percent.)

Hope that helps,

William