So I try to solve this question: there is a row of 7 elements, and I need to find the probability of element b and a (which are some arbitrary elements) are separated by at least 1 of the other elements present in the row.

So since we had covered similar questions on the lessons I knew that it was more efficient to find the probability of them being together and subtract it rather then probability of them being separated because of scenarios when one element is on the edge of the row.

Since when they are together they form a block of 2 elements they they will have six different positions the block can take. I apply factorial to this 6 to account for movement of elements outside the pair, and multiply by 2 to account for the arrangement of element a & b within the pair.

Then I divide it by factorial of 7: $$ P_{together} = \frac{2\times{6!}}{7!} = \frac{2}{7} $$ $$ P_{separated} = 1 - \frac{2}{7} = \frac{5}{7} $$

But the answer to the question states that the answers for $$P_{\,together}$$ and $$ P_{\,separarted}$$ should be switched: Screenshot

Finally please take into account that, while this resource is meant for IB curriculum students, I ensured I have a legal means to use it for revision as my school acquired the license for it.

  • $\begingroup$ I`ve edited my issue to clarify it. @lulu $\endgroup$ Commented May 27, 2021 at 14:11
  • 1
    $\begingroup$ That would appear to be a typo. Clearly, it is more likely that they be separated than that they be together. $\endgroup$
    – lulu
    Commented May 27, 2021 at 14:11


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