Probability of choosing same multiple

Question

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. — Max Freiburghaus, Mar 13 '17 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. — Ian L, Mar 13 '17 at 22:42
"Betty Wont randomly selects ..." reads "Betty will randomly ... ", I suppose. — GNUSupporter 8964民主女神地下教會, Mar 13 '17 at 22:44
@lulu From the edit history of this question, I think OP means $\frac{3}{17\times19}$ is "obviously incorrect". — GNUSupporter 8964民主女神地下教會, Mar 13 '17 at 22:48
@lulu I was writing from my old work before I tried a second time at the problem. I meant to write 10 options instead of 19. — Ian L, Mar 13 '17 at 22:50
Ok...but the current version says that $\frac 3{170}$ is obviously incorrect and I don't see why. It looks right to me. Am I missing something? — lulu, Mar 13 '17 at 22:51
@GNUSupporter "Betty Wont" is her name. It is an attempt at humour. "Will E. Pickett / Betty Wont", get it? — Graham Kemp, Mar 14 '17 at 00:08
@GrahamKemp Yes. The author typically uses whimsical names just for humor. — Ian L, Mar 14 '17 at 00:13

score 1 · Accepted Answer · answered Mar 14 '17 at 00:31

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.

Probability of choosing same multiple

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