# S1 Question

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#1
http://www.ocr.org.uk/Images/136156-...atistics-1.pdf
Need help with 7iib) :-(
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6 years ago
#2
This question becomes a lot easier when you treat them both as one 'item'. Therefore, we have 5 people and 1 item. Then its just working out the number of combinations that contain the item, over the total number of combinations.

Considering when they are part of the group, the number of combinations can be defined as 5C3, as there are 3 spots left (two have been taken by Jill and Jo).

The number of combinations when they aren't in a group is 5C5. Therefore the probability is 5C3/(5C3 * 5C5). This comes out to 10/11.
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