binomial distributions

Question

Find the probability of having 2,3, or 4 successes in five trials of a binomial experiment in which the probability of success is 40% Round to the nearest tenth of a percent.

Answer 1

P(x) = the desired probability

n = Number of trials = 5

x = Number of successes = 2, 3, or 4

p = probability of success = 40% or 0.40

q = probability of failure = 1 - p = 1 - 0.40 = 0.60

For 2 success, x = 2

P(2) = 5!/[(5-2)!·2!]·(0.40)²·(0.60)^5-2 = 5!/[(5-2)!·2!]·(0.40)²·(0.60)³ = 5!/(3!·2!)·(0.16)·(0.216) = 10·(0.16)·(0.216)

= 0.3456

The probability of 2 successes is 34.6%

For 3 and 4 successes, x = 3 and x = 4. You would run the same calculation 2 more times changing all the x values to the new number.

Answer 2

The general formula is _NC_X(p)^X(q)^(N-X)

where N = the # of trials

X = the # of successes

p = probability of success

q = probability of failure (1-p)

For 2 successes ₅C₂(.4)²(.6)^(5-2) = (10)(.16)(.216) = .3456 or 34.6%

You do the others the same way!