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.
The formula is:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
In this problem:
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:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 0) = h(0,7,2,3) = \frac{C_{3,0}C_{4,2}}{C_{7,2}} = 0.2857[/tex]
[tex]P(X = 2) = h(2,7,2,3) = \frac{C_{3,2}C_{4,0}}{C_{7,2}} = 0.1429[/tex]
Hence:
[tex]p = P(X = 0) + P(X = 2) = 0.2857 + 0.1429 = 0.4286[/tex]
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