Cars and Trucks!

By mathhelpguy on Apr 17, 2007 | Reply

3C2 * .05^2 * .95^1 = .007125

By Merlyn on Apr 20, 2007 | Reply

Let X be the number of cars sold. X has the binomial distribution with n = 3 trials and success probability p = 0.05

In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, …, n
P[X = x] = 0 for any other value of x.

The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.
Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.

X ~ Binomial( n , p )

the mean of the binomial distribution is n * p = 0.15
the variance of the binomial distribution is n * p * (1 - p) = 0.1425
the standard deviation is the square root of the variance = ? ( n * p * (1 - p)) = 0.3774917

The Probability Mass Function, PMF,
f(X) = P(X = x) is:

P( X = 0 ) = 0.857375
P( X = 1 ) = 0.135375
P( X = 2 ) = 0.007125 ? answer
P( X = 3 ) = 0.000125