# Thread: Binomial ,with replacement

1. ## Binomial ,with replacement

found this question on a website , it gives the answer as 0.27
but it doesn't show how the author arrived at the answer
ive never dealed with replacement before

A bag contains 5 yellow beads and 5 blue beads (all identical apart from the colour).
7 beads are drawn randomly from the bag, each being replaced before the next is drawn.
Calculate the probability that
Exactly 4 of the beads drawn are yellow

7! / 4! 3! = 35 * multiplied by 0.5 seven times= 0.2734

I dont know if the formula used is correct
or i simply got lucky and matched authors answer ?

2. I would like to know why you multiplied 0.5 seven times.

did you arrive to your answer by using this formula:

$\dbinom{n}{x}p^{x}(1-p)^{n-x}$??

3. I would like to know why you multiplied 0.5 seven times.

because the chance of drawing a yellow bead or blue is 1 in 2

did you arrive to your answer by using this formula:
that formula you used is a bit advanced for me!!
im a standard calculator guy !!

4. okay. let me ask you something before I answer your question.

what would you have done if the question was same except that you had 6 yellow and 4 blue beads?

5. so the new question would be
A bag contains 6 yellow beads and 4 blue beads (all identical apart from the colour).
7 beads are drawn randomly from the bag, each being replaced before the next is drawn.
Calculate the probability that
Exactly 4 of the beads drawn are yellow

7! / 4! 3! * 0.6 *0.6 *0.6 * 0.6 * 0.4 * 0.4 * 0.4 = 0.2903

6. Right.

So your answers are correct in both the cases. And the formula you are using is correct indeed!

Note that you are using the same formula I have written in post #2.

Here's an explanation(your original question):

since there are 5 yellow and 5 blue beads,

the probability of obtaining a yellow bead $p=5/10=0.5$

the probability of obtaining a yellow bead $(1-p)=1-0.5=0.5$

Total number of balls = 7.

Let X be the number of yellow balls drawn. then,

$P(X=x)=\dbinom{n}{x}p^{x}(1-p)^{n-x}$ $\mbox{where}\dbinom{n}{x}=\dfrac{n!}{x!(n-x)!}$

the probability of 4 yellow balls being drawn is:

$P(X=4)=\dbinom{7}{4}\times(0.5)^{4}\times(0.5)^{7-4}$

$P(X=4)=\dfrac{7!}{4!\times 3!} \times (0.5)^4\times(0.5)^3$

$=\dfrac{7!}{4!\times 3!} \times (0.5)^7= 0.27$

7. tx , found the question on a website was trying for 2 hours to find a formula to get the same answer as the
author !!