I found that the probability of Jack and John to be in different groups is 5/9, and thought that the probability of them being in the same group is 1- 5/9 = 4/9. My mathematics book says that it is C8,3/C10,5=2/9.
[Math] 10 people are divided in two equal groups. What is the probability that two given people (Jack and John) are at the same group
probability
Best Answer
You’re right; the book is wrong. It’s true that there are $\binom83$ ways to choose $3$ other people to be in a group with Jack and John, so there are $\binom83$ groups of $5$ that include both of them. However, there are only $\frac12\binom{10}5$ ways to split the $10$ into two groups of $5$, not $\binom{10}5$: $\binom{10}5$ counts each split twice, once for each of the two groups of $5$ making it up. Thus, the denominator should have been $\frac12\binom{10}5$, and the final result would then have been the correct $\frac49$.
Your approach is much simpler, however: just observe that in any split Jack has an equal chance of being any of the other $4$ people in John’s group or the $5$ people in the other group, so he has $4$ chances in $9$ of being in John’s group.