Question

Of all the Harry Potter books purchased in a recent year, about 60% were purchased for readers 14yrs or older. If 12 Harry Potter fans who bought the book that year are surveyed, find the following probabilities.
a) At least 5 of them are 14 or older.
b) Exactly nine of the are 14 or older
c) Less than 3 of them are 14 or older.

Answer 1

Using the binomial distribution, it is found that:

a) 0.9427 = 94.27% probability that at least 5 of them are 14 or older.

b) 0.1419 = 14.19% probability that exactly nine of the are 14 or older.

c) 0.0028 = 0.28% probability that less than 3 of them are 14 or older.

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For each book, there are only two possible outcomes. Either it was purchased by readers 14yrs or older, or it was not. The probability of a book being purchased by a reader 14yrs or older in independent of any other book, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of a success on a single trial.

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• 60% were purchased by readers 14yrs or older, thus $p = 0.6$
• Sample of 12 books, thus $n = 12$

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Item a:

The probability is:

$P(X \geq 5) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.6)^{0}.(0.4)^{12} \approx 0$

$P(X = 1) = C_{12,1}.(0.6)^{1}.(0.4)^{11} = 0.0003$

$P(X = 2) = C_{12,2}.(0.6)^{2}.(0.4)^{10} = 0.0025$

$P(X = 3) = C_{12,3}.(0.6)^{3}.(0.4)^{9} = 0.0125$

$P(X = 4) = C_{12,4}.(0.6)^{4}.(0.4)^{8} = 0.0420$

Then

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0 + 0.0003 + 0.0025 + 0.0125 + 0.0420 = 0.0573$

$P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0573 = 0.9427$

0.9427 = 94.27% probability that at least 5 of them are 14 or older.

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Item b:

This is P(X = 9), thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{12,9}.(0.6)^{9}.(0.4)^{3} = 0.1419$

0.1419 = 14.19% probability that exactly nine of the are 14 or older.

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Item c:

This is:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Using the probabilities from item a:

$P(X < 3) = 0 + 0.0003 + 0.0025 = 0.0028$

0.0028 = 0.28% probability that less than 3 of them are 14 or older.

A similar problem is given at brainly.com/question/15557838