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    Can someone please explain the answers to these questions. exam board is OCR MEI
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    Attachment 529009529011Attachment 529009529011Attachment 529009529011529000[attach]5.29009529011529e+23[/attach]
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    for part i)

    you have to choose 3 prize winners from 20. then decide how many ways they may be allocated the 3 different prizes...
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    (Original post by the bear)
    for part i)

    you have to choose 3 prize winners from 20. then decide how many ways they may be allocated the 3 different prizes...
    yeah but I thought you would use nCr for that?
    20C3
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    (Original post by kennyboy69.5)
    yeah but I thought you would use nCr for that?
    20C3
    you need that, but each of those choices can be arranged in a certain number of ways....
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    (Original post by the bear)
    you need that, but each of those choices can be arranged in a certain number of ways....
    oooh right so 20C3 x 3! =6840
    thanks what about the other question?
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    Both halves of Question 6 can be done without the nCr method.
    For 6i, imagine the prizes are given out in order of the English one first, French second, and Spanish last. Since in this instance only one student can win each prize, and after that prize can't win the others, think about process of elimination:
    Any of the 20 students could win the English prize, so we have 20 possible winners. Let's say, for example, a boy called Aaron wins this prize.
    Next, for the French prize, since Aaron has already won the English prize, it can't go to him, therefore 19 of the 20 students can win it. Turns out that a girl called Bethany wins the French prize.
    The final Spanish prize, can be won by anybody except Aaron and Bethany, who already have prizes. So 18 can win this one.
    Don't worry about the order; 20 have the chance to win the first prize, 19 can win the second, and 18 can win the third.
    The mathematical formula using the factorial rule (total number of possibilities / (factorial of desired possibilities x factorial of difference between total and desired possibilities) can be cancelled down to simply give us this:


    For the second question, 6ii, remember that each of the three English, French and Spanish prizes can be given away to any of the 20 students. Forget limitations for now, and just cube 20 to give you 8000.
    But this therefore includes the possibility that any one student can win all the prizes, when we don't want that. Therefore we subtract 20 (as there are 20 possible cases that any one student can win all three prizes).
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    For the other question (Question 4), for part ii,the first magnet he gets can be any of the 5 available so thats 5/5 or 1, but when he opens the next packet, the probability he got the first one will be 1/5; hence the answer is 1 x 1/5.

    Part ii, is kinda similar however the first try, it doesn't matter what he gets first hence it's 5/5, but on the second packet he can have any of the 4 remaining magnets he hasn't yet gotten so it's 4/5, and then the 3rd packet - he has two different ones already but he still has 3 to find so it's 3/5 and so on until the 5th packet where it would be the last remaining magnet hence that probability would be 1/5.

    hence the answer would be 5/5 x 4/5 x 3/5 x 2/5 x 1/5

    For Part iii, well, you just do 1 minus the answer you got from part ii. This is because part ii basically is the probability that he gets all types of fridge magnets in 5 packets, hence if he would be need more than 5 packets to get a complete set, it would be 1 - P(5 Packets).
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    Thanks so much for your explanations! totally understand this now
 
 
 
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