Question

There is a bag filled with 3 blue and 4 red marbles.
A marble is taken at random from the bag, the colour is noted and then it is not replaced.
Another marble is taken at random.
What is the probability of getting 2 of the same colour?
There is a bag filled with 3 blue and 4 red marbles.
A marble is taken at random from the bag, the colour is noted and then it is not replaced.
Another marble is taken at random.
What is the probability of getting 2 of the same colour?

Answer 1

Using the hypergeometric distribution, it is found that there is a 0.4286 = 42.86% probability of getting 2 of the same colour.

The marbles are chosen without replacement, hence the hypergeometric distribution is used to solve this question.

What is the hypergeometric distribution formula?

The formula is:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• 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.

In this problem:

• There is a total of 3 + 4 = 7 marbles, hence N = 7.

• Of those, 3 are blue, hence k = 3.

• 2 marbles will be taken, hence n = 2.

The probability of getting 2 of the same colour is the sum of P(X = 0), which is both red, with P(X = 2), which is both blue, then:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,7,2,3) = \frac{C_{3,0}C_{4,2}}{C_{7,2}} = 0.2857$

$P(X = 2) = h(2,7,2,3) = \frac{C_{3,2}C_{4,0}}{C_{7,2}} = 0.1429$

Hence:

$p = P(X = 0) + P(X = 2) = 0.2857 + 0.1429 = 0.4286$

0.4286 = 42.86% probability of getting 2 of the same colour.

You can learn more about the hypergeometric distribution at brainly.com/question/4818951