Question

Suppose that you have an urn with 500 balls, 100 of which are red and 400 are black. (a) You sample 10 balls at random with replacement. What is the probability that at least 2 of them are red? (b) You sample 10 balls at random without replacement. What is the probability that none of them are red? (expression only)

Answer 1

a) The probability that a randomly selected ball is red in each trial $= \frac{100}{500} = 0.2$

Let X denotes the number of red balls in a sample 10 balls selected at random with replacement

X ~ Binomial(n = 10, p = 0.2)

The probability mass function of X is

$P(X= x) = \binom{10}{x}*0.2^x*(1-0.2)^{10-x},x=0,1,2,...,10$

Now,

The probability that at least 2 of them are red

$=P(X\geq 2)$

$= 1 - P(X<2)$

$=1-P(X\leq 1)$

$=1 -\sum_{x=0}^{1} \binom{10}{x}*0.2^x*(1-0.2)^{10-x}$

$= 1 - 0.375810$

$= 0.624190$

b) Let Y denotes the number of red balls in a sample 10 balls at random without replacement.

Y ~ Hypergeometric(N = 500, M = 100, n = 10)

The probability mass function of Y is

$P(Y =y) = \frac{\binom{100}{y}*\binom{400}{10-y}}{\binom{500}{10}},y=0,1,2,...,10$

Now,

The probability that none of them are red

$= 1 - P(Y =0)$

$=1 - \frac{\binom{100}{0}*\binom{400}{10-0}}{\binom{500}{10}}$