A coin is slightly bent, and as a result the probability of a head is 0.52. Suppose that you toss the coin four times. Use the Binomial formula to find the probability of 3 or more heads
1 - P(0)+P(1)+P(2)
the Ps in my head are all one term ... so i spose they should technically be wrapped in ( )
Can anyone help with the binomial question. I think I have the correct answer but need to fully understand
the hard part is not that hard ... its just using the binomial formula. What do you have as the binomial formula?
i spose P(3)+P(4) is simpler to assess, less Ps. I was thinking this was a lot more trials for some reason
\[\binom{4}{3}p^3q^1+\binom{4}{4}p^4q^0\] such that p is a success and q is (1-p), a failure
Yes, that's what i have i ended up with 4/16 which is .25
rechk your mathing ... 4/3 = 4, and 4/4 = 1 are the most common errors http://www.wolframalpha.com/input/?i=%5Cbinom%7B4%7D%7B3%7D%28.52%29%5E3%281-.52%29%5E1%2B%5Cbinom%7B4%7D%7B4%7D%28.52%29%5E4%281-.52%29%5E0
that and getting the probabilities mixed up ...
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