Thread: Marbles and Probability

A bag of marbles contains 10% brown, 10% red, 20% yellow, 20% blue, 20% orange and 20% green.

Given the information about the distribution, now consider you pick 4 marbles in a row.

What is the probability that:

a. all 4 are blue?
b. at least one is red?
c. the forth chosen is the first to be brown?

I don't think this is enough information to answer this. Suppose there are 100 marbles.

The probability of selecting a blue marble on the first pick is $\frac{1}{5}$

The probability of selecting a blue marble on the second pick is $\frac{19}{99}$

3rd pick $\frac{18}{98}$, 4th $\frac{17}{97}$

The product of these is 0.00123558.

Now suppose there are 1000 marbles.

These probabilities become $\frac{1}{5}, \frac{199}{999}, \frac{198}{998}, \frac{197}{997}$

and the product of these is 0.00156179

in the limit as $N \to \infty$ you'd expect that the probability of selecting 4 blues becomes

$\left(\frac{1}{5}\right)^4=\frac{1}{625}=0.0016$