probabilities?

Question

A box contains 35 red and 5 black discs. a disc is selected at random and its colour is noted. the disc is then replace in the box

in eight selections find the probability that a black disc is selected

exactly once and atleast once

Answer 1

Each selection has p=(1/8) chance of black, and (1-p)=(7/8) chance of red.

The combination function C(n,m) is defined as
C(n,m) = n! / (m! * (n-m)!)

In 8 trials, the chance of getting exactly n blacks is
P(n) = C(8,n) * p^n * (1-p)^(8-n)

The sum of P(n) over n=0 to n=8 is 1.0; you have to get some number of black disks in the 8 trials, so the sum of probabilities is 100%.

Therefore, the chance of exactly one black is
P(1)

and the chance of at least one black is
P(1) + P(2) + ... + P(8)
= 1 - P(0)

I'll let you calculate the actual numbers.

added afterwards -- a comment on Been There's answer:
He calculated the chance that the first disk will be black and the next 7 will be red, but what you wanted was the chance that exactly one would be black. With 8 selections, the black disk could be picked on the first selection, the second, etc. There are 8 different ways for this to happen, so the chance of exactly one black disk is not 4.9% but 8 times this (39.27%). The above formulas give you this answer.

Answer 2

Let X be the number of black discs that are selected. X has the binomial distribution with 8 trials and success probability p = 5/40.

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.

this is found 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, to be more accurate, the binomial is the sum of n independent and identically distributed Bernoulli trials.

the mean of the binomial distribution is n * p
the variance of the binomial distribution is n * p * (1 - p)
the standard deviation is the square root of the variance.

P(X = 0 ) = 0.3436089
P(X = 1 ) = 0.3926959
P(X = 2 ) = 0.1963480
P(X = 3 ) = 0.05609941
P(X = 4 ) = 0.01001775
P(X = 5 ) = 0.001144886
P(X = 6 ) = 8.177757e-05
P(X = 7 ) = 3.33786e-06
P(X = 8 ) = 5.960464e-08

P( X ≥ 1) = 1 - P(X = 0) = 1 - 0.3436089 = 0.6563911

Answer 3

Assuming that it does not matter which selection results in a black disk, here's a path:

You know that probability of selecting exactly one black disk on a single turn is 5/40 or 1/8. Again assuming that the turn in which the black disk is selected does not matter, and you will take 8 turns from the complete set of 35 red and 5 black disks, we may also assume that the trial are independent of each other. These two points are important, as the assumptions will drive the rules of evaluation.

Once you define the assumptions, the math is quite simple.

x = pull exactly one black disk in 8 trials.
P(x) = (1/8)(7/8)(7/8)(7/8)(7/8)(7/8)(7/8)(7/8)
= (0.125)(0.875)(0.875)(0.875)(0.875)(0.875)(0.875)(0.875)
= (0.125)(0.3927)
= 0.049

The short version , you have a 4.9% probability of pulling exactly one black disk throughout the course of the eight trials.