Anne worked at a firm with 30 people. There were 20 workers and 10 managers. A party planning committee was formed, but Anne was not invited to participate. When she went to the HR director to find out why she wasn't on the committee, the HR director told her the five people were selected at random. 3 workers and 2 managers had to be on the committee, but the 3 workers were picked out of a hat from the 20. The same procedure was used to select the managers.
Anne went to a lawyer (she was considering suing! How could they not pick her for the party planning committee, it was obviously scandalous!), who consulted with a statistician. The statistician told the lawyer to forget it. The odds against Anne being picked for the committee were ...
BTW, this is the post that was supposed to go out yesterday. Blogger was down for most of the day (I don't know if you noticed or not).
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20% or .2
ReplyDeleteAssume, Anne is a worker then probability of she getting selected is 3/20 = 0.15
Assume, she is a manager then probability is 2/10 = .2
In best case scenario, the probability is 20%.
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I think it's a bit more complicated than that. The odds of Anne being picked equals:
ReplyDelete(the odds of her being a worker x the odds of a worker being picked) + (the odds of her being a manager x the odds of a manager being picked)
In numbers that is:
(2/3 x 3/20) + (1/3 x 2/10) = 1/10 + 2/30 = 1/6
That's 0.16666666
She never stood a chance!
Ann had 0 chance, she was not amoung the 30 workers.
ReplyDeleteI really like that last answer! Totally thinking outside the question.
ReplyDeleteguessthelogo and reallyfatbloke have laid out great explanations, so I won't expand on them. I'll leave it to others to say if we need to calculate the odds of her being a manager or worker.
In a stats class, sure. But in real life, I would hope the lawyer would have known what her position was! ;-)