# A bag contains 6 red balls and 7 green balls. You plan to select 5 balls at random. What is the probability of selecting (without replacement) 5 green balls?

Dec 12, 2016

Probability of taking 5 green balls $= \frac{21}{1287} = 0.016$

#### Explanation:

Given -

red balls $= 6$
green balls$= 7$

Total $= 6 + 7 = 13$

Five balls can be taken from 13 balls in ${C}_{5}^{13}$ ways.

C_5^13=(n!)/((n-r)!r!)=(13!)/((13-5)!*5!)
=(13xx12xx11xx10xx9xxcancel(8!))/(cancel(8! )xx5xx4xx3xx2xx1)=154440/120=1287

Five green balls can be taken from 7 green balls in ${C}_{5}^{7}$ ways.

C_5^7=(n!)/((n-r)!r!)=(7!)/((7-5)!*5!)
=(7xx6xxcancel(5!))/((2xx1)xxcancel(5!))=42/2=21

Probability of taking 5 green balls $= \frac{{C}_{5}^{13}}{{C}_{5}^{7}} = \frac{21}{1287} = 0.016$