Question 590674: you have a book collection that consists of 20 horror novels, 15 novels,and 25 mystery novels. you randomly pick 4 books to read during a long trip. what is the probability that you pick at least one book of each type?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! a = 20 = number of horror novels.
b = 15 = number of regular novels.
c = 25 = number of mystery novels.
total number of books is equal to 20 + 15 + 25 = 60.
you pick 4 books at random.
you need to pick at least 1 of each type.
3 things have to happen.
you do not pick 0 type a or you do not pick 0 type b or you do not pick 0 type c.
the probability of picking 0 type a is equal to 40/60 * 39/59 * 38/58 * 37/57 = P(40,4)/P(60,4).
the probability of picking 0 type b is equal to 45/60 * 44/59 * 43/58 * 42/57 = P(45,4)/P(60,4).
the probability of picking 0 type c is equal to 35/60 * 34/59 * 33/58 * 32/47 = P(35,4)/P(60,4).
the combined probability of picking 0 type a or picking 0 type b or picking 0 type c is equal to:
(P(40,4) + P(45,4) + P(35,4)) / (P(60,4)
this comes out to be equal to .600336317
this is the probability that you will get {0 type a or 0 type b or 0 type c}.
the probability that you will not get {0 type a or 0 type b or 0 type c} is equal to 1 minus .600336317 which is equal to .399663683.
not getting 0 of each type means that you will get at least 1 of each type.
this is difficult to confirm that the formulas work because the possible combinations are 81 and each one of those possible combinations has its own probability, the sum of which must be equal to 1.
it's easier to show if you have 2 choices of books and are drawing 3 books.
in this case, there are 8 possible combinations so it's easier to show the probability of each combination and to show that the sum of the probabilities is equal to 1.
it's also easier to then confirm that the formula of 1 minus the probability of {0 type a or 0 type b} provides the correct answer.
those details are shown below:
number of type a 40
number of type b 20
total 60
summary probability formula used
p(0 type a) = 0.033313852 equals 20/60*19/59*18/58 ***
p(1 type a and 2 type b) = 0.222092344 equals 3*40/60*20/59*19/58
p(2 type a and 1 type b) = 0.455873758 equals 3*40/60*39/59*20/58
p(0 type b) = 0.288720047 equals 40/60*39/59*38/58 ***
total probability 1
details probability formula used
aaa 0.288720047 equals 40/60*39/59*38/58 ***
aab 0.151957919 equals 40/60*39/59*20/58
aba 0.151957919 equals 40/60*20/59*39/58
abb 0.074030781 equals 40/60*20/59*19/58
baa 0.151957919 equals 20/60*40/59*39/58
bab 0.074030781 equals 20/60*40/59*19/58
bba 0.074030781 equals 20/60*19/59*40/58
bbb 0.033313852 equals 20/60*19/59*18/58 ***
total probability 1
the summary shows you that the probability of getting 0 type a is the same as the probability of getting 3 type b.
it also shows you that the probability of getting 0 type b is the same as the probability of getting 3 type a.
the probability of getting 3 type a or 3 type b is shown in the details.
the probability of getting at least 1 of each type is equal to 1 minus the probability of getting 0 type a or 0 type b.
the same procedure is used in the larger numbers where we picked 4 books from a selection of 3.
finding the number of 0 type a combination was more complicated in this case because there were 16 combinations where 0 type a were picked and 16 combinations where 0 type b were picked and 16 combinations where 0 type c were picked.
for example, the possible combinations where type a books were not picked are:
bbbb
bbbc
bbcb
bbcc
bcbb
bcbc
bccb
bccc
cbbb
cbbc
cbcb
cbcc
ccbb
ccbc
cccb
cccc
this made it difficult to show the details because the individual calculations became onerous to say the least.
it was much easier to reduce the number of possibilities in order to show you how it worked.
all of these combinations, however, involved 1 minus the probability of getting 0 type a or 0 type b or 0 type c.
the same calculations in the simpler example produced the same result, i.e. 1 minus probability of 0 type a minus probability of 0 type b gave the result of at the probability of least 1 of each type.
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