Question #dded5

1 Answer
Jul 24, 2016

"No. of novel"=3

"No. of books of poem"=4

"No. of dictionary"=1

"Total No. of books"=3+4+1=8

(i)
3 books can be chosen from total 8 books in ""^8C_3 ways.Keeping dictionary a fixed choice other 2 books can be chosen from rest 7 books in ""^7C_2 ways.

Hence the probability that among the three chosen book one is the dictionary,is (""^7C_2)/(""^8C_3)=21/56=3/8

(ii)
Amongst the 3 books chosen there wii be 1 novel and 2 books of poem.This can be done in ""^3C_1 xx""^4C_2 ways.
So the probability of this choice will be (""^3C_1 xx""^4C_2) /(""^8C_3)=(3*6)/56=9/28