Digit d is randomly selected from the set {1,2,3,4,5,6,7,8,9}. Without replacement of d, another digit, e, is selected. What is the probability that the two-digit number de is a multiple of 3?

Question

Digit d is randomly selected from the set {1,2,3,4,5,6,7,8,9}. Without replacement of d, another digit, e, is selected. What is the probability that the two-digit number de is a multiple of 3?

1 Answer

Impact of this question

5478 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-11T09:55:30

The probability that a two-digit number formed from the set of digits ${1, 2, 3, 4, 5, 6, 7, 8, 9}$ ${1, 2, 3, 4, 5, 6, 7, 8, 9}$ without replacement happens to be a multiple of $3$ $3$ is $\frac{1}{3}$ $1/3$ .

Explanation:

Recall that for any multiple of 3, the sum of its digits is also divisible by 3. Now we may reframe this question to be, "what is the probability that $d + e$ $d + e$ is divisible by $3$ $3$ ?"

Let's say $d$ $d$ is $1$ $1$ . Out of the remaining $9$ $9$ options, $2$ $2$ , $5$ $5$ , and $8$ $8$ will all give us a multiple of $3$ $3$ .

Similary, $2$ $2$ has $1$ $1$ , $4$ $4$ , and $7$ $7$ .

This pattern holds for all integers from 1-9 with the exception of $3$ $3$ , $6$ $6$ , and $9$ $9$ which all have only $2$ $2$ options as you cannot add the same number.

So, six values for $d$ $d$ have $3$ $3$ working values for $e$ $e$ , and three values have $2$ $2$ working values. Calculating the number of ways the sum of $d$ $d$ and $e$ $e$ is a multiple of 3:

$6 \cdot 3 + 3 \cdot 2$ $6 * 3 + 3 * 2$

$= 18 + 6$ $= 18 + 6$

$= 24$ $= 24$

And finally, take this as a fraction of the total number of permutations, which is the number of different numbers you can pick for $d$ $d$ , times the number of ways you can pick $e$ $e$ . This is $9 \cdot 8$ $9*8$ as we do not replace $d$ $d$ .

$\frac{24}{9 \cdot 8}$ $24/(9*8)$

$= \frac{24}{72}$ $=24/72$

$= \frac{1}{3}$ $=1/3$

So, the probability that a two-digit number formed from the set of digits ${1, 2, 3, 4, 5, 6, 7, 8, 9}$ ${1, 2, 3, 4, 5, 6, 7, 8, 9}$ without replacement happens to be a multiple of $3$ $3$ is $\frac{1}{3}$ $1/3$ .

Science

Math

Humanities

... and beyond

Digit d is randomly selected from the set {1,2,3,4,5,6,7,8,9}. Without replacement of d, another digit, e, is selected. What is the probability that the two-digit number de is a multiple of 3?

1 Answer

Explanation:

Related questions

Impact of this question