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    Our maths teacher said in order to do this question - just draw the whole tree diagram and add all the probabilities. The problem with this is that it would take far too long. And I read you can use combination to solve this. Problem is my permutation and combinations is really bad so I can't really grasp the concept here. Can someone explain it? (I already have the answer but I just want explanations for the steps including combination)
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    each probability is the number of ways of choosing n number of girls/ number of ways of choosing 4 people from a group of 8.

    For example; for there to be 3 girls there has to be 1 boy. So the number of ways of choosing 3 girls is  5c3 \times 3c1

    All the possible combinations of 4 from 8 are  8c4

    So the number of ways of choosing 3 girls is  \frac{5c3 \times 3c1}{8c4} = \frac{3}{7}
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    (Original post by NotNotBatman)
    each probability is the number of ways of choosing n number of girls/ number of ways of choosing 4 people from a group of 8.

    For example; for there to be 3 girls there has to be 1 boy. So the number of ways of choosing 3 girls is  4c3 \times 4c1

    All the possible combinations of 4 from 8 are  8c4

    So the number of ways of choosing 3 girls is  \frac{4c3 \times 4c1}{8c4} = \frac{8}{35}
    ...but there are 5 girls and 3 boys to choose from, so isn't it

     \frac{5c3 \times 3c1}{8c4} = \frac{3}{7} ?
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    (Original post by old_engineer)
    ...but there are 5 girls and 3 boys to choose from, so isn't it

     \frac{5c3 \times 3c1}{8c4} = \frac{3}{7} ?
    Yes, that's what I meant (read it incorrectly) . I'll edit the post to reflect this. The mark scheme should have the same answer.
 
 
 
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