Re: Monty at it again
I don't think I can explain this any clearer .
The probability 1 or 2 gets selected on the first draw is 2/10 , agree ?
Given that 1 or 2 has been selected , the other number gets removed from the vault so that there are 8 numbers left . The probability it's B should be 2/8 since he has two good choices out of 8 . Agree ?
However , B may hit 3 or 4 on the first draw rather than the second with probability 2/10 . Agree ?
Given that B hits a 3 or a 4 , we would remove the other number so that there are 8 numbers left . The probability A hits a 1 or a 2 from 8 possible choices is 2/8 . Agree ?
Now we add 2/10*2/8 + 2/10*2/8 = 0.1
Do the other examples the same way .
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