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?

1 Answer
Dec 12, 2016

Probability of taking 5 green balls #=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 #=(C_5^13)/(C_5^7)=21/1287=0.016#