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    1) Each of the digits 3, 4, 5, 6 and 7 are written on a separate piece of card. Four of the cards are picked at random .

    How many odd numbers smaller than 5000 are possible?

    -------------------------
    There are four spaces, the last digit can have 3 outcomes , and the first digit can have 2 outcomes. So there are 2 spaces left in the middle, that need to be filled from the 3 cards left, so :

    3P2 * 2 * 3 = 36

    The answer meant to be 30


    2) Three letters are now selected at random, one at a time, from the 8 letters of the word COMPUTER,
    and are placed in order in a line.
    (iii) Find the probability that the 3 letters form the word TOP.
    ------------
    I drew out 3 spaces, the first would be 8C1 x 7C1 x 6C1 = 336 . I'm not sure how to calculate the Probability from this ? Probability is what I want / possible outcomes, 336 is what I want , what would the outcomes be ? The answer is 1/336 though
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    The outcomes of the first digit and last digit are not independent. For example, if the first digit is 3, then that eliminates 3 from the three outcomes for the last digit.
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    I'm confused on how to do the calculation as a result of that.

    So there would be 2 choices for the first still, but this time 2 for the last now instead of 3 . Now there are 2 choices for the remaining spaces with 3 choices to choose from:
    2*3P2*2= 24
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    I assume there is some conditional probability involved, but I suck at even the most basic probability
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    Yeah I keep getting it wrong =/ . Thanks for the help. Added one more question that I got wrong .
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    1) If the first card is fixed as 3, then for the next two cards we may select from the 4,6 and an odd (5 or 7)...3P2 ways.
    The final card has a choice of two odds. Hence number of ways is 3P2x2 = 12

    If the first card is fixed as a 4, then for the next two cards we may select from an even card and two odds...3P2 ways.
    The final card has a choice of three odds. Hence number of ways is 3P2x3 = 18

    12+18 =30 ways in total.

    2) The no. of ways of selecting 3 letters from 8 is 8P3 = 336.
    So the probability of forming TOP is 1/336.
    If you think of it as a tree diagram it's 1/8x1/7x1/6 = 1/336
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    (Original post by vc94)
    1) If the first card is fixed as 3, then for the next two cards we may select from the 4,6 and an odd (5 or 7)...3P2 ways.
    The final card has a choice of two odds. Hence number of ways is 3P2x2 = 12

    If the first card is fixed as a 4, then for the next two cards we may select from an even card and two odds...3P2 ways.
    The final card has a choice of three odds. Hence number of ways is 3P2x3 = 18

    12+18 =30 ways in total.

    2) The no. of ways of selecting 3 letters from 8 is 8P3 = 336.
    So the probability of forming TOP is 1/336.
    If you think of it as a tree diagram it's 1/8x1/7x1/6 = 1/336

    Thanks for the help

    Just not sure why its 1/336 , is there a set formula that says 1/N=P ? Without the use of a tree diagram ?
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    Total no. of possible 3 letter words is 336. Each one is equally likely, so probability of each outcome is 1/336.
 
 
 
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