Question

A beg contain 14 identical balls. 4 of which are red, 5 black and 5 white. Six balls are from the beg. Find the probability that (i) 3 are red (ii) at least two are white.(iii) 4 black balls. *

Answer 1

Probability of success P of any event = $\frac {Total\ number\ of\ favorable\ outcomes\ 'F'}{Total\ number\ of\ possible\ outcomes\ 'T'}$

Total number of outcomes in selecting 6 balls out of 14 balls by using the below formula,

n=14, r=6

$T=\ \frac {n!}{(n-r)!*r!} = \frac {14!}{(14-6)!*6!} = 3003$

Answer 1)

Since we require 3 red balls and remaining 3 balls of any color the total number of favorable outcomes is given by,

$F=4C3*10C3=\frac {4!}{(4-3)!*3!}*\frac {10!}{(10-3)!*3!}=480$

Here we have considered the selection of remaining 3 non-red balls as a group of 10.

Hence the probability of success for getting exactly 3 red balls from the bag is given by,

$P=\frac {F}{T}=\frac {480}{3003} = 0.15984$

Answer 2)

Here we require atleast 2 white colored balls which means we can have 2 or 3 or 4 or all 5 white balls while selecting a total of 6 balls whose total number of favorable outcomes is given by,

$F= 5C2*9C4+5C3*9C3+5C4*9C2+5C5*9C1=2289$

Hence the probability of getting atleast 2 white balls in the selection is given by,

$P=\frac {F}{T}=\frac {2289}{3003}=0.7622377$

Answer 3)

Since we require exactly 4 black balls the total number of favorable outcome is given by,

$F=5C4*9C2=180$

Hence the probability of getting exactly 4 Black balls is given by,

$P=\frac {180}{3003}=0.05994$