Difference between revisions of "2023 WSMO Accuracy Round Problems/Problem 4"
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| + | The probability the person in front of Bob is his friend is <math>\frac{3}{9}</math>. The probability that the person behind Bob is his friend given that the persin in front of Bob is his friend <math>\frac{2}{8}</math>. So, our answer is <cmath>\left(\frac{3}{9}\right)\left(\frac28\right)=\frac6{72}=\frac1{12}\implies1+12 = \boxed{13}.</cmath> | ||
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| + | ~pinkpig | ||
Latest revision as of 11:46, 13 September 2025
Problem
Bob and his 3 friends are standing in a line of 10 people. Given that Bob is not on either end of the line, then the probability the person in front and behind Bob are both his friends is 
for relatively prime positive integers 
and 
Find 
Solution
The probability the person in front of Bob is his friend is 
. The probability that the person behind Bob is his friend given that the persin in front of Bob is his friend 
. So, our answer is ![\[\left(\frac{3}{9}\right)\left(\frac28\right)=\frac6{72}=\frac1{12}\implies1+12 = \boxed{13}.\]](http://latex.artofproblemsolving.com/1/3/a/13a066e0132e023b263cfa00c168f7271d1389cb.png)
~pinkpig
