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Which one of these is more likely to occur?

A.The sequence H-T-H-T-T-H is more likely to happen than the sequence H-H-H-T-T-T.

B.The sequence H-T-H-T-T-H is less likely to happen than the sequence H-H-H-T-T-T.

C.The sequence H-T-H-T-T-H is equally likely to happen as the sequence H-H-H-T-T-T.

Jojo
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    Site users generally prefer that you show a bit of work for questions like this, otherwise they just feel like they're just being used to do your homework for you rather than actually help you learn. – 727 Sep 10 '15 at 21:42
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    @LJL This is me preparing for interviews actually, so I'm going to have to understand the question one way or another. I personally thought the answer was C as I feel both sequences are equally likely. – Jojo Sep 10 '15 at 21:44
  • For independent tosses of a fair coin, each of the $2^6$ possible outcomes is equally likely. The probability of getting exactly 3 heads and 3 tails in any order is ${6 \choose 3}/2^6.$ Is there context to this problem assuring you that the coin is fair and tosses are independent? – BruceET Sep 10 '15 at 21:48
  • @BruceTrumbo I think we have to assume the coin is fair and independent. – Jojo Sep 10 '15 at 21:48
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    Then my Comment and the Answer by 'user-whatever' are correct. – BruceET Sep 10 '15 at 21:51
  • For independent tosses of a coin, fair or not, the events are equally likely. – André Nicolas Sep 10 '15 at 22:29
  • This question probably refers to the mean times before various motives appear, in which case the answer is that these are not equal. – Did Sep 11 '15 at 05:55

1 Answers1

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Assuming that you have an unbiased coin ... the probabilities of all answers given would actually be the same.

The probability of flipping a head=probability of flipping a tail= 1/2

Since you flip the coin six times in all of the given scenarios... the probability of all answers is simply $(1/2)^6$

Therefore, all of the given sequences are equally likely to occur. Answer is C.

user264985
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