Lindsey randomly selects two distinct days each week to go rock-climbing at the gym. Jess randomly selects three distinct days each week to climb at the same gym. What is the probability that Lindsey and Jess will both be at the gym on the same day at least once in any given week? Express your answer as a common fraction.
I thought the answer should be 1-36/2401
$P(Meet >=1) = 1-P(Meet=0)$
The probability of meeting 0 times is just the probability of any arrangment of meeting times the number of ways we can choose 5 unique days out of the week.
$P(Meet=0) = 3/7 * 2/7 *1/7 *2/1 * 1/7 * C^7_5$
$=6*6/7^4$
$=36/2401$
However, the correct answer is 5/7. What is wrong with my methodology?
Best Answer
Jess picks $3$ days. We want the probability that Lindsay picks a day that Jess has picked. It's easier to compute the complement. The probability that Lindsay picks both days from the two that Jess hasn't picked is $${{4\choose2}\over{7\choose2}}={{4\cdot3}\over{7\cdot6}}=\frac27$$ so the probability we seek is $$1-\frac27=\frac57$$
EDIT
When I woke up this morning, I realized how you were attempting to do the problem, or at least I think I do. You want to say, for any five days of the week, compute the probability that Jess and Lindsay pick those five days, and then multiply by the number of ways to choose five days. That will work, but you've computed the probability incorrectly.
Suppose the five days have been chosen. Jess can choose three of these days in ${5\choose3}$ ways and he can choose three days of the week in ${7\choose3}$ ways, so the probability of success is $${{5\choose3}\over{7\choose3}}$$ Then Lindsay must choose the remaining two days, and he has ${7\choose2}$ choices, so the probability that he chooses the remaining two days is $${1\over{7\choose2}}$$ Now, we must add these probabilities up for every possible choice of five days and there are ${7\choose5}={7\choose2}$ such choices. Altogether, we have a probability of $$ {{5\choose3}\over{7\choose3}}{1\over{7\choose2}}{7\choose2}= {{5\choose3}\over{7\choose3}}= {{5\cdot4\cdot3}\over{7\cdot6\cdot5}}={2\over7}$$
I think that the basic flaw in your approach is trying to do the problem too quickly. It seems to me that you are writing down calculations without stopping to think what they mean, or why they should be correct. This is very tempting, I know, but you should resist the temptation. Try to get into the habit of writing down, in words, what you are doing, and why it is correct. If you don't, you may find that when you go back and look at the calculation later, you may find that you can't understand it, even if it is correct, or that you can't figure out why you thought it was correct, if it was wrong. This has happened to me many times.
Also, of course, if you don't do this, it is harder to communicate with others. Look how long it's taken me to figure out the gist of your calculations. Somebody smarter than I would have seen it it a lot sooner no doubt, and I may have been sleepy yesterday, but even so, it shouldn't have taken so long.
I hopes this helps.