Question

An urn contains 7 blue and 8 green balls. You remove 3 balls from the urn without replacement. What is the probability that at least 2 out of the 3 balls are green.

Answer 1

Answer)

Here we have 7 blue and 8 green balls = 15

We need to take out 3 balls

Probability is given by favorable/total

Toyal = 15c3

Ncr = n!/(r!*(n-r)!)

Where n! = n*n-1*n-2*n-3*n-4*n-5*....till 1

For example

5! = 5*4! = 5*4*3! = 5*4*3*2! = 5*4*3*2*1

So, 15c3 = 455

Favorable = at least 2 green

Here we have two cases

First case = 2 green and 1 blue

8c2*7c1 = 28*7 = 196

Second case = all green

= 8c3

= 56

Favorable = 196+56 = 252

Probability = 252/455 = 0.55384615384