Question

A box contains 3 red marbles, 5 white marbles and 7 blue marbles, 2 balls are drawn at random. Calculate probability? a, There is 1 red ball out of the 2 balls drawn b, There is at least 1 red ball out of 2 balls drawn

Answer 1

a) 0.3429 = 34.29% probability there is 1 red ball out of the 2 balls drawn.

b) 0.3715 = 37.15% probability that there is at least 1 red ball out of 2 balls drawn

Explanation:

The balls are drawn without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

In this question:

3 + 5 + 7 = 15 balls, which means that
`N = 15`

3 red, which means that
`k = 3`

2 drawn, which means that
`n = 2`

a) Probability there is 1 red ball out of the 2 balls drawn

This is
`P(X = 1)`

So

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 1) = h(1,15,2,3) = (C_(3,1)*C_(12,1))/(C_(15,2)) = 0.3429`

0.3429 = 34.29% probability there is 1 red ball out of the 2 balls drawn.

b) Probability there is at least 1 red ball out of 2 balls drawn

This is:

`P(X \geq 1) = P(X = 1) + P(X = 2)`

So

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 1) = h(1,15,2,3) = (C_(3,1)*C_(12,1))/(C_(15,2)) = 0.3429`

`P(X = 2) = h(2,15,2,3) = (C_(3,2)*C_(12,0))/(C_(15,2)) = 0.0286`

`P(X \geq 1) = P(X = 1) + P(X = 2) = 0.3429 + 0.0286 = 0.3715`

0.3715 = 37.15% probability that there is at least 1 red ball out of 2 balls drawn