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Difference between revisions of "1985 AJHSME Problems/Problem 22"

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There are <math>10-2=8</math> possibilities for the first digit in total, and only <math>1</math> that works, so the probability the number begins with <math>9</math> is <math>\frac{1}{8}</math>
 
There are <math>10-2=8</math> possibilities for the first digit in total, and only <math>1</math> that works, so the probability the number begins with <math>9</math> is <math>\frac{1}{8}</math>
  
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There are <math>10</math> possibilities for the last digit, and only <math>1</math> that works <math>(0)</math>, so the probability the number ends with <math>0</math> is <math>\frac{1}{10}</math>
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Out of the <math>10</math> possibilities for the last digit, only <math>1</math> works: <math>(0)</math>, so the probability the number ends with <math>0</math> is <math>\frac{1}{10}</math>
  
 
Since these are [[Independent event|independent events]], the probability both happens is <cmath>\frac{1}{8}\cdot \frac{1}{10}=\frac{1}{80}</cmath>
 
Since these are [[Independent event|independent events]], the probability both happens is <cmath>\frac{1}{8}\cdot \frac{1}{10}=\frac{1}{80}</cmath>

Revision as of 23:35, 9 September 2020

Problem

Assume every 7-digit whole number is a possible telephone number except those that begin with $0$ or $1$. What fraction of telephone numbers begin with $9$ and end with $0$?

$\text{(A)}\ \frac{1}{63} \qquad \text{(B)}\ \frac{1}{80} \qquad \text{(C)}\ \frac{1}{81} \qquad \text{(D)}\ \frac{1}{90} \qquad \text{(E)}\ \frac{1}{100}$

Note: All telephone numbers are 7-digit whole numbers.

Solution 1

An equivalent problem is finding the probability that a randomly selected telephone number begins with $9$ and ends with $0$.

There are $10-2=8$ possibilities for the first digit in total, and only $1$ that works, so the probability the number begins with $9$ is $\frac{1}{8}$

Out of the $10$ possibilities for the last digit, only $1$ works: $(0)$, so the probability the number ends with $0$ is $\frac{1}{10}$

Since these are independent events, the probability both happens is \[\frac{1}{8}\cdot \frac{1}{10}=\frac{1}{80}\]

$\boxed{\text{B}}$

Solution 2

The fraction is simply the number of $7$-digit phone numbers with the restrictions applied divided by the total number of phone numbers. Let $a$ denote the numerator, and $b$ denote the denominator. Let's first work on finding $b$, the total number.

For a regular digit, there are $10$ possible choices to make: $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, or $9$. The only digit that is not regular is the first one, which prohibits $0$ and $1$ from taking place, resulting in $8$ possible choices to make for that first digit. Since each digit is independent of one another, we multiply the number of choices for each digit, resulting in $8 * 10 * 10 * 10 * 10 * 10 * 10$, or $8 * 10 ^ 6$ possible total phone numbers ($b$).

Now that we have the denominator, the only unknown remaining is $b$. To solve for $a$, let's use the same method as we did for the denominator. For the first digit, there is only one possible value: $9$. For the last digit, there is only one possible value: $0$. However, the rest of the five digits are "free" (meaning they can be any one of $10$ choices). Thus $a = 1 * 10 * 10 * 10 * 10 * 10 * 1$, or $10^5$ possible phone numbers with restrictions.

The fraction $\frac{a}{b}$ is the same as $\frac{10^5}{8 * 10^6}$, which reduces to $\boxed{\text{B}}$.

(~thelinguist46295)

See Also

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions


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