Is over there a method to fix the trouble considering that the probability of getting a head is 1/2 and also then calculating $.5^4$ and also multiplying $.5^4$ through 4 as there space 4 ways that this might occur? $\begingroup$ my book says the answer is 5/16. $$4nCr 3 + 4 ncR 4 / (2^4)$$ $2^4$ originates from that fact that there room $2^4$ outcomes in the sample space. How is my systems wrong? $\endgroup$
The probability the at least three heads deserve to be found by$$\sum_k = 3^4\binom4k.5^k.5^4-k = \frac516$$The factor being is us have four coins and we desire to select 3 or much more heads. Therefore, we sum the the binomial distribution for 4 choose 3 and also 4 choose 4 through probability of a fair coin so $p = q = 0.5$.

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The binomial circulation is$$\binomnkp^kq^n -k$$where $q = 1 - p$. The price is 5/16 due to the fact that in complete there space 16 possibilities when tossing a coin 4 times. I.e. $2\cdot2\cdot2\cdot2 = 2^4$.

And the possibilities of obtaining atleast three head is 5 the is us can obtain HHHT, HHTH, HTHH, THHH, and HHHH  Thanks because that contributing solution to beer-selection.comematics stack Exchange!

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