An urn contains 4 green , 5 blue ,2 red and 3yellow marbles .
If four marbles are drawn at randon , what is the probability that two are blue and two are red ?

Options

A. `(10)/(1001)`
B. `(9)/(14)`
C. `(17)/(364).`
D. `(2)/(7)`

Answer

A

Explanation

According to question ,
`n(S) =""^(14)C_(4)=(14!)/((14-4)!4!)=(14!)/(10!4!)`
`=(14xx13xx12xx11)/(4xx3xx2xx1)=1001[:' ""^(n)C_(r) =(n!)/((n-r)!r!)]`
` and n( E) =""^(5)C_(2)xx""^(2)C_(2)=(5!)/((5-2)!2!)xx(2!)/((2-2)!2!)`
`=(5xx4)/(2xx1)xx(2xx1)/(1xx2xx1)=10`
`implies` Required probability `=(n( E))/(n(S))=(10)/(1001)`