Binomial Distribution Question: Ribbons in a Bag

Question

A bag contains a very large number of ribbons. One quarter of the ribbons are yellow and the rest are blue. Ten ribbons are selected at random from the bag.

a) Find the expected number of yellow ribbons selected.

b) Find the most likely number of yellow ribbons selected.

I have done a), but do not know how to approach b) (answer is supposed to be 2).

Let random variable $X$ be the number of yellow ribbons selected. $n = 10$, $p = 1/4$, $q = 1- p = 3/4$

a) $E(X) = np = 10 \left(\frac{1}{4}\right) = 2.5$

Answer 1

First assume that the sample follows a binomial distribution (even though the probability of success changes, it is negligible due to the large sample size).

Then, define your random variable: Ex. Let X be the number of Yellow Ribbons selected.

Next, create a binomial probability distribution function with the probabilities for each outcome P(X=x) where x equals 0, 1, 2, ... 10.)

The outcome x with the largest probability is the modal value (the value of the random variable that occurs most frequently).

Cheers!

Answer 2

You've calculated the mean for a binomial distribution for (a). $\mathsf E(X)= np = 2.5$   That's a good approximation for sampling from a very large (but unknown) population size.

So you'll want the mode of a binomial distribution for (b). That is: $\lfloor (n+1)p\rfloor = 2$

Or you could do it the hard way and find the maxima of the probability mass function; either analytically or by hunt-and-peek.

$$\mathsf P(X=x) \mathop{:=} \dbinom n x p^x (1-p)^{n-x}$$