![]() ![]() Therefore, by the principle of inclusion-exclusion,Īnd the number of sequences of $\ n\ $ tosses in which both heads and tail occur at least twice in a row is thereforeįor $\ n=100\ $ Wolfram alpha gives the number of such sequences as The intersection $\ A\cap B\ $ of these sets is the set of sequences containing no successive heads or tails. The complement is the union of the set $\ A\ $ of sequences containing no successive heads with the set $\ B\ $ of sequences containing no successive tails. The easiest way of getting the number of sequences of $\ n\ $ tosses in which both heads and tail occur at least twice in a row seems to be to use the principle of inclusion-exclusion to obtain the number of elements in its complement. The number of sequences of tosses with no consecutive heads or tails is always $2$, because heads and tails must alternate, so the whole sequence is completely determined by the result of the first toss. If there are three heads in the sequence of five coin tosses, the only possibility is that the sequence is HTHTH. If there are four or five heads in the sequence of five coin tosses, at least two heads must be consecutive. Your proposed answer of $13/32$ is correct. 'COIN-FLIP OUTCOME' is a 15 letter Phrase starting with C and ending with E Crossword answers for COIN-FLIP OUTCOME Synonyms for TAILS 4 letter words coat 6 letter words blazer chases formal jacket show 9 more results Thanks for visiting The Crossword Solver 'Coin-flip outcome'. What is the probability that heads never occurs twice in a row? ![]() There are two possibilities for each of the five tosses of the coin, so there are $2^5 = 32$ possible outcomes in your sample space, as you found.
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