A book contains 92 pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is 9?

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Solution

Number of pages of the book = 92
Which are from 1 to 92
Number of possible outcomes = 92
$∴$ Number of pages whose sum of its page is 9
=10 i.e. 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
$∴$ P(E)= $\frac{10}{92} = \frac{5}{46}$

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A book contains 92 pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is 9?

Number of pages of the book = 92 Which are from 1 to 92 Number of possible outcomes = 92 ∴ Number of pages whose sum of its page is 9 =10 i.e. 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 ∴ P(E)= 1092=546

Number of pages of the book = 92
Which are from 1 to 92
Number of possible outcomes = 92
$∴$ Number of pages whose sum of its page is 9
=10 i.e. 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
$∴$ P(E)= $\frac{10}{92} = \frac{5}{46}$