Problem of the Day


Perry buys a book that has pages numbered from 1 to 980. He then selects one of the pages at random. What is the probability that the number of the page he selects is divisible by 7, given that it is divisible by 3?

A. \displaystyle \frac{1}{21}

B. \displaystyle \frac{1}{7}

C. \displaystyle \frac{23}{163}

D. \displaystyle \frac{23}{490}

E. \displaystyle \frac{29}{171}

Reveal Answer

Answer

[latexpage]

C. $\displaystyle \frac{23}{163}$

See the Solution

Solution

[latexpage]

This is a conditional probability problem. We’re asked to find the probability that the page number is divisible by 7,¬†given¬†that it is divisible by 3. The set of possible outcomes consists of all multiples of 3 from 1 to 980. To determine the probability in question, we need to determine how many of these numbers are also divisible by 7. Because these numbers will be divisible by both 3 and 7, it follows that they will also be divisible by 21. In other words, we are trying to find how many numbers from 1 to 980 are divisible by 21. As a general rule, to determine how many multiples of n¬†there are from 1 to x, divide x¬†by n¬†and disregard the remainder. In this case, we need to divide 980 by 3 and 980 by 21. The first calculation tells us that there are 326 multiples of 3 from 1 to 980 and the second calculation tells us that there are 46 multiples of 21. So, 46 of the multiples of 3 from 1 to 980 are also divisible by 7. Therefore, the probability of the page number being divisible by 7 given that it is divisible by 3 is $\displaystyle \frac{46}{326}=\frac{23}{163}$.


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