Question

A box consists of red and blue marbles. Five marbles are selected randomly, and the chance of selecting a red marble is 40%. Assume that once a marble is selected, that it is placed back in the box. Use the Binomial distribution to compute the probabilities.
a. All 5 marbles selected are red.
b. At most 2 of the marbles selected are red.
c. Exactly 3 of the marbles selected are red.

I know that for these types of problems, you need to figure out what is n, X, p, and q...
I know that p = 0.4 and q = 0.6
I also think that X changes. For instance, in a. it's 5?
I need help with n!

Answer 1

In each part of the question X is the number of red marbles in 5 draws. The number of draws is 5, therefore n = 5. Your results for p and q are correct.
Do you know the formula for the binomial distribution?

Answer 2

Yes, the binomial distribution formula is: P(X) = nCx (p)^x (q)n-x

So for a., would I solve it like this?
P(5) = 5C5 (.4)^5 (.6)^0 = 0.01024

Answer 3

Yes, your solution to a is correct.

Answer 4

and for b.,
P(1) = 5C1 (.4)^1 (.6)^4 = 0.2592
P(2) = 5C2 (.4)^2 (.6)^3 = 0.3456

P(at most 2) = 0.2592 + 0.3456 = 0.6048

If this is also correct, then I think I understand the question :)

Answer 5

For b you also need to add the probability that X = 0 to the sum of the probabilities
P(X = 1) and P(X = 2).

Answer 6

Ooohh, okay! I forgot about that! Thank you very much for your help ^-^d

Answer 7

You're welcome :)