I have 1000 books, 400 of which are math books. a) if i randomly and without rep

Question

I have 1000 books, 400 of which are math books.

a) if i randomly and without replacement, select 60 books.Find the exact probability that either 23 or 24 of the books i choose are math books.

b) suppose that i randomly select a book, determine if it a math book and then put it back in the collection, repeat this 60 times. find the exact probablity that i will select either 23 or 24 maths books

c) Use a normal approximation to estimate the answer b

(Please show me all the procedures of how you get to the answer thanks)

Explanation / Answer

Binomial Distribution
PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
a.
P( X = 23 ) = ( 60 23 ) * ( 0.4^23) * ( 1 - 0.4 )^37
= 0.1018
P( X = 24 ) = ( 60 24 ) * ( 0.4^24) * ( 1 - 0.4 )^36
= 0.1047
P(X=23 OR X=24) = 0.1018 + 0.1047 = 0.2065
b.
suppose that i randomly select a book, determine if it a math book = 0.4
and for next attempt we are putting it back to the book collection and drawing a math book, it will be again .4
probablity that i will select either 23 or 24 maths = (0.4)^23 + (0.4)^24 = (0.4)^23 [ 1 + 0.4 ] = 0.000000000985162
c.
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 23) = (23-24)/3.7947
= -1/3.7947= -0.2635
= P ( Z <-0.2635) From Standard Normal Table
= 0.3961
P(X > 24) = (24-24)/3.7947
= 0/3.7947 = 0
= P ( Z >0) From Standard Normal Table
= 0.5
P( X < 23 OR X > 24) = 0.3961+0.5 = 0.8961                  

P ( X = 23 OR 24 ) = 1 - 0.8961 = 0.1039

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