A bit string of length 10 is generated by flipping a coin 10 times (Head =1, Tail =0). If the coin is biased so that H is 3 times more likely to come than T, what is the probability that the string contains at most 3 zeroes
This will be a binomial distribution with parameters:
n = 10, p = P(Tails) = 1/4
q = P(Heads) = 1 - 1/4 = 3/4
Hence,
P(Atleast 3 zeros)
= P(0) + P(1) + P(2) + P(3)
= 10C0 (1/4)0 (3/4)10 + 10C1 (1/4)1 (3/4)9 + 10C2 (1/4)2 (3/4)8 + 10C3 (1/4)3 (3/4)7
= 0.7759
A bit string of length 10 is generated by flipping a coin 10 times (Head =1,...
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