# Question dded5

Jul 24, 2016

$\text{No. of novel} = 3$

$\text{No. of books of poem} = 4$

$\text{No. of dictionary} = 1$

$\text{Total No. of books} = 3 + 4 + 1 = 8$

(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 $\left({\text{^7C_2)/(}}^{8} {C}_{3}\right) = \frac{21}{56} = \frac{3}{8}$

(ii)
Amongst the 3 books chosen there wii be 1 novel and 2 books of poem.This can be done in ${\text{^3C_1 xx}}^{4} {C}_{2}$ ways.
So the probability of this choice will be $\left({\text{^3C_1 xx""^4C_2) /(}}^{8} {C}_{3}\right) = \frac{3 \cdot 6}{56} = \frac{9}{28}$