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 This topic has 5 replies, 2 voices, and was last updated 3 years, 4 months ago by Ayush.

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VishalGuest
Hello,
On the 9/20/16 meeting, we solved the following problem in class:
(CCCMC Mock AMC 8) A group of 20 students is being divided up into 2 groups of 10. Find the probability that Olivia is in a group with Oliver if Olivia and Oliver are part of the group of 20 students.Is there a second solution to this problem using combinations?
adminKeymaster
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Hey Vishal,That’s an excellent question that was also asked during the lecture.
But to reiterate, here is another possible argument you could have used:Without regard to order, we split the $20$ students into 2 groups $A$ and $B$. If both Olivia and Oliver are in group A, then the remaining $18$ students can be arranged into $2$ groups in $\dbinom{18}{10}$ ways since $10$ of them must go to group $B$ and $8$ of them must go to group A randomly. Similarly, we also have $\dbinom{18}{10}$ arrangements if both Oliver and Olivia are in group B. Thus, our desired solution is $\frac{2\dbinom{18}{10}}{\dbinom{20}{10}}=\frac{9}{19}$.
PRATYUSHGuestWhy does it matter to us that in which group they are, because the arrangement will still be the same
AyushGuestYou are correct it doesn’t matter what group they are in, but if you look at the numerator of the fraction we obtain at the end, there is a factor of 2 that is being multiplied by the combination. That factor of 2 results from the fact that they could be placed in either group.
nimalanGuestGet it but what if they are put in different groups, there would be nine spots in each group so isn’t it different from the combination of (18 10) when it id (19 9). please explain if my explanation is wrong or tell me if you do not understand me
AyushGuestGood question Nimalan, the problem statement says that they must be in the same group. But you are correct, if they were in separate groups then we do 18 choose 9.

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