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Help needed with probability calculation

Discussion in 'Mathematics' started by geoff1954, Feb 19, 2020.

  1. geoff1954

    geoff1954 New commenter

    I know it ought to be easy but I can't remember how to do this calculation. Please help. It is a real life problem, not made up for the sake of it. I suspect that some sex discrimination has gone on.

    There are twelve applicants for a job, three men and nine women. are six places available. . What is the probability that six women, and no men, are selected for the job (assuming equal probability of all applicants.)
  2. Corvuscorax

    Corvuscorax Star commenter

  3. Stiltskin

    Stiltskin Star commenter

    Probability of a woman being picked for the first job 9/12
    Probability of a woman being picked for the second job 8/11
    Probability of a woman being picked for the third job 7/10
    Probability of a woman being picked for the forth job 6/9
    Probability of a woman being picked for the fifth job 5/8
    Probability of a woman being picked for the final job 4/7

    Probability of all women getting the jobs roughly 9/100 or 9%
    bombaysapphire likes this.
  4. geoff1954

    geoff1954 New commenter

    Thanks for the response, but surely it will be much less than 7/8?
  5. geoff1954

    geoff1954 New commenter

    That sounds about right. Thanks. So suspicious, but not grounds for action.
  6. Corvuscorax

    Corvuscorax Star commenter

    I mean against

    each candidate has a 50% chance of getting a job. Each man has a 50% chance of getting a job
  7. catherinedavid

    catherinedavid New commenter

    If no men are selected in this event, then the chance of all women being picked would surely be 3/4
  8. Stiltskin

    Stiltskin Star commenter

    There are 924 permutations of the candidates (n=12, r=6)
    84 of those permutations are all women so 84/924 that all jobs go to women (or 9%)
    There is a 100% chance that at least three women will get a job but again only 9% chance all three men would be chosen.
  9. Corvuscorax

    Corvuscorax Star commenter

    no, it would be 100%
    Stiltskin likes this.
  10. Stiltskin

    Stiltskin Star commenter

    Theres about a 41% chance there would only be one man chosen and 41% that two men were chosen.

    It is possible then that some unconscious bias was involved in selecting who to recruit.
    anniestgermain likes this.
  11. GeordieKC

    GeordieKC Occasional commenter

    Probability only tells you how likely an event is to occur. If the probability is greater than zero then that event can occur.

    You cannot conclude that a low probability means the event should not occur unless there is bias involved - the probability of a particular person winning the lottery is very small but it still happens
    Lara mfl 05 likes this.
  12. bombaysapphire

    bombaysapphire Star commenter

    I agree with the 9% calculation but I would be concerned about assuming that all candidates are equally qualified. If just one of the women was clearly outstanding it would mean you were picking for 5 places from 3 men and 5 women and the probability of selecting all women increases to 12%.
    Lara mfl 05 likes this.
  13. gainly

    gainly Lead commenter

    I agree with @Stiltskin's calculation. although it should of course be combination, not permutation. I'm not sure where 7/8 came from.
    bombaysapphire and Stiltskin like this.
  14. briancant

    briancant Occasional commenter

    (9! x 6!)/(12! x 3!)=0.0909090....

    Bet if there was 9 men and 3 women it would have been a bit different.
    afterdark likes this.
  15. dave153

    dave153 New commenter

  16. grumbleweed

    grumbleweed Lead commenter

    Bit rusty but I made it 1/11 so 9.1%.

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