A fair six-sided die is thrown n times. Let Pn be the probability that the total number of times of obtaining a "six" in the n throws is an odd number.
Show that Pn = (2/3)Pn-1 + (1/6)
Please teach me with explanation...Actually I dont quite understand what is the question is asking about....Thank you !!!
Probability
2012-01-01 10:02 pm
回答 (1)
2012-01-01 10:27 pm
✔ 最佳答案
After n throws, if the no. of times of obtaining a 'six' is odd, then there are two prossibilities.I The n th throw obtaining 'six' and so the no. of times of obtaining a 'six' in the previous n - 1 throws should be even
2 The n th throw does not obtain 'six' and so the no. of times of obtaining a 'six' in the previous n - 1 throws should be odd
So, we can deduce that
P(n) = (1/6)(1 - P(n - 1)) + (5/6)(P(n))
P(n) = (2/3)P(n - 1) + 1/6
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