Question #dded5

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Question #dded5

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Answer 1 · 2016-07-24T15:55:44

$No. of novel = 3$ $"No. of novel"=3$

$No. of books of poem = 4$ $"No. of books of poem"=4$

$No. of dictionary = 1$ $"No. of dictionary"=1$

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

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

Hence the probability that among the three chosen book one is the dictionary,is $\frac{^{7} C_{2}}{^{8} C_{3}} = \frac{21}{56} = \frac{3}{8}$ $(""^7C_2)/(""^8C_3)=21/56=3/8$

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

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Question #dded5

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