Question

There are four books. In each book all the pages are numbered (excluding the covers). The first has 28 pages, the second 82 pages, third 98 pages and fourth 148 pages. If the book is selected at random then what is the probability that it opens at page 84?

Answer 1

Here we will calculate the probability that a book selected at random opens at page 84 keeping in mind that as only one book is to be selected and the order of the books is not important. We will subtract the books which have lesser pages and use the concept of combination on other two books to find probability.

• We have combination formula as $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$, where we choose $r\ ]items from \[n$ items.

Complete step-by-step answer:
First book contains 28 pages, so the probability that the first book selected at random opens at page 84 is zero.
Second book contains 82 pages, so the probability that the second book selected at random opens at page 84 is also zero.
Third book contains 98 pages, so the probability that third book selected at random opens at page 84 is calculated as $P(Page\,84) = \dfrac{{P(F)}}{{P(T)}}$, where P(F) is the probability of the favourable outcome and P(T) is the probability of the total number of outcomes. Here favourable outcome is one because it is just one page.
$P(Third\\_Page\,84) = \dfrac{1}{{98}}$for the third book
Similarly, fourth book contains 148 pages, so the probability that fourth book selected at random opens at page 84 is calculated as,
$P(Fourth\\_Page\,84) = \dfrac{1}{{148}}$for the fourth book
The probability of selecting 1 out of the 4 books is calculated using combination as $P(book){ = ^4}{C_1}$ since only one book is to be selected and order of the book is not important.
The two of the four books can give favourable outcomes; the probability of the favourable outcomes comes out of the total outcomes $P(total){ = ^4}{C_2}$.
The probability that book selected at random opens at page 84 is given as
$= \dfrac{{P(Book) \times P(Third\\_Page\,84) \times P(Fourth\\_Page\,84)}}{{P(total)}}$ ... (1)
$P\left( {Third\\_Page{\text{ }}84} \right),{\text{ }}P\left( {Fourth\\_Page{\text{ }}84} \right)$ is the probability of Each outcome.
$P\left( {Book} \right)$ is the probability of selecting each outcome from the total outcomes
$P\left( {total} \right)$ is the probability of the total outcomes.
Substituting the values in the formula we get
$= \dfrac{{^4{C_1} \times \dfrac{1}{{98}} \times \dfrac{1}{{148}}}}{{^4{C_2}}}$ ... (2)
We know from the formula of combination is $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
So $^4{C_1} = \dfrac{{4!}}{{(4 - 1)!1!}} = \dfrac{{4!}}{{3!}}$
$= \dfrac{{4 \times 3!}}{{3!}} = 4$ { since $n! = n \times (n - 1)!$}
And $^4{C_2} = \dfrac{{4!}}{{(4 - 2)!2!}} = \dfrac{{4!}}{{2!.2!}}$
$= \dfrac{{4 \times 3 \times 2!}}{{2 \times 2!}} = 6$
Substituting the values in the formula.
$\dfrac{{^4{C_1} \times \dfrac{1}{{98}} \times \dfrac{1}{{148}}}}{{^4{C_2}}} = \dfrac{4}{6} \times \dfrac{1}{{98}} \times \dfrac{1}{{148}}$
$= \dfrac{1}{{21756}}$ … (3)
The probability that a book selected at random opens at page 84 is $\dfrac{1}{{21756}}$.

Note: Students are likely to make mistakes when they write they include the first two books in the choosing process of total probability as well, keep in mind only that book can open page 84 which has a total number of pages greater than 84 pages.