# Probability of choosing same multiple

Will E. Pikett randomly selects an odd integer less than $100$ that is a multiple of $3$. Betty Wont randomly selects an odd integer less than $100$ that is a multiple of $5$. What is the probability that they selected the same number?

My approach:

The number of odd integers that are less than $100$ and a multiple of $3$ is $17$. As for odd multiples of $5$, that is $10$.

There are $3$ factors in common: $15$, $45$, and $75$.

So the probability of choosing one of these three factors for Will is $\frac{3}{17}$. The probability that Betty will choose the same number is $\frac{1}{10}$ (Betty could have chosen first I suppose).

So the probability that they both chose the same factor is $\frac{3}{170}$, but obviously I'm incorrect. Where in my work did I make an erroneous decision, and what is the result of choosing such a decision. Thanks.

• you've just miscounted the ODD multiples of 5. Everything else is correctly argued from there. Commented Mar 13, 2017 at 22:40
• Forgive me. I was typing off my old work. My new work did account for $10$ odd multiples of $5$. I still got the wrong answer apparently. Commented Mar 13, 2017 at 22:42
• "Betty Wont randomly selects ..." reads "Betty will randomly ... ", I suppose. Commented Mar 13, 2017 at 22:44
• Why do you say that this is "obviously incorrect"?
– lulu
Commented Mar 13, 2017 at 22:46
• @lulu From the edit history of this question, I think OP means $\frac{3}{17\times19}$ is "obviously incorrect". Commented Mar 13, 2017 at 22:48

That is okay.   It was correctly reasoned, you merely had difficulty counting / identifying the numbers.

Your approach has been to use: \begin{align}\mathsf P(B=W) ~&=~ \mathsf P(W\in\{15,45,75\})~\mathsf P(B=W\mid W\in\{15,45,75\}) \\[1ex] &=~ \frac 3{17}\cdot\frac{1}{10}\end{align}

Now that the comments have lead you to the proper counts of odd numbers less than 100 that are multiples of three, and same of five, that is entirely correct.