Posted by KeKe on Sunday, September 16, 2012 at 1:28pm.
To get it right, make sure you distinguish between the first and second throw. If you do that, you'll see that there are 36 distinct outcomes: (1, 1), (1, 2), (1, 3), ... (6, 4), (6, 5) and (6, 6). They're all equally likely with probability 1/36, so how many of them total more than 5? We could just enuerate them all - or better, work out how many total five or less (because there are fewer of them), and then subtract the probability from 1. These are: (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2) and (4, 1). There are 10 of them, so I reckon the answer ought to be 1 - (10/36) = 0.72. Do you agree?
Thank you so much David Q/R.
You're welcome!
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