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bagirrra123 [75]
3 years ago
6

Calculate the probability of landing at most 1 head when a 0.6-biased coin is tossed

Mathematics
1 answer:
asambeis [7]3 years ago
8 0

Answer:

\frac{2^{n-1}}{5^n}(2n+3)

Step-by-step explanation:

We have given the coin is 0.6 biased means p=0.6=\frac{6}{10}=\frac{3}{5}

q =1-p=1-\frac{3}{5}=\frac{2}{5}

We have to find the at most one head means sum of probability of 0 head and 1 head

So p=p(x=0)+p(x=1)

From binomial distribution the probability is given by p=^nC_rp^rq^{n-r}

So probability p=^nC_0\frac{3}{5}^0\frac{2}{5}^{n-0}+^nC_1\frac{3}{5}^1\frac{2}{5}^{n-1}=\frac{2^{n-1}}{5^n}(2n+3)

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