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I'm just looking for a confirmation. Textbook answer for this problem is $\frac{1}{18}$ but I'm getting $\frac{1}{20}$...

You have 7 tickets to a baseball game. Three tickets are in the left field stands, 2 are behind home plate and 2 are near first base. Two tickets are randomly given to your sister. You notice that one of the tickets is not behind home plate. Calculate the probability that both tickets are near first base.

$P(A|B)=\cfrac{\frac{1}{21}}{\frac{20}{21}}=\cfrac{1}{20}$

Am I missing something? Is there a faster way to solve these problems than doing a probability tree or listing the possible outcomes?

hawks10
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    Is your assumption that exactly one or at least one of the two tickets is not near Home? – lulu Mar 22 '22 at 21:26
  • @lulu the way I see it, we only know that one is not near home, we can't assume anything about the other...? – hawks10 Mar 22 '22 at 21:31
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    So, it's at least one then? The probability that at least one is not near home is $1-\frac 1{21}=\frac {20}{21}$. The probability that both are near first is $\frac 1{21}$, so I agree with your calculation. – lulu Mar 22 '22 at 21:33
  • Note: there are $10$ combinations with exactly one not near home, making the probability of that $\frac {10}{21}$. The numerator is unchanged, making the answer $\frac 1{10}$ in that case. No guess where the $\frac 1{18}$ came from. – lulu Mar 22 '22 at 21:35
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    @lulu Thanks for your help! I notice that if the condition was one of the tickets is not in the stands, the answer would be $\frac{1}{18}$... it's the only thing I can see... – hawks10 Mar 22 '22 at 21:43
  • Oh, good catch. That seems as likely as anything. – lulu Mar 22 '22 at 22:19

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