There are 9 books on the shelf. We can pick up 3 book in
"\\displaystyle C(9,3) = \\frac{9!}{3!6!} = \\frac{7 \\cdot 8 \\cdot 9}{3 \\cdot 2}=84" ways.
a) We can pick up 3 novels out 5 available in
"\\displaystyle C(5,3) = \\frac{5!}{3!2!} = \\frac{5 \\cdot 4}{2}=10"
So, the probability of "all books selected are novels" is
"P(A) = \\frac{10}{84}= \\frac{5}{42}"
b) We can pick up 2 novels and 1 book of poems in
"\\displaystyle C(5,2) \\cdot C(3,1) = \\frac{5!}{2!3!} \\cdot \\frac{3!}{2!1!}= 10 \\cdot 3=30"
The probability of event "2 novels and 1 book of poems are selected" is
"P(A) = \\frac{30}{84}= \\frac{5}{14}"
Answer: a) 5/42, b) 5/14
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