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A fair coin is tossed 10 times. What is the probability that the result is 5 consecutive heads followed by 5 consecutive tails?

I'm thinking 10! should be the amount of combinations that can be made. There is only 1 combination that has 5 heads followed by 5 tails. I'm not sure if 1/10! is the answer or if I need to include something about tails and heads both having 1/2 probability.

Thanks for the help!

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    $\begingroup$ This is just one outcome out of $2^{10}$, so the probability ( assuming independence) is $1/2^{10}$. $\endgroup$ – user99680 Feb 14 '14 at 1:52
  • $\begingroup$ Heads and tails having equal probability is all you need to include... $\endgroup$ – DJohnM Feb 14 '14 at 1:55
  • $\begingroup$ Thanks for the comments. That makes a lot more sense. $\endgroup$ – Liliana Feb 14 '14 at 2:07
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There are $2^{10}$ possible outcomes when flipping a coin 10 times, and, like you said, there is only one way for the event " five consecutive heads followed by five consecutive tails" to happen (to be pedantic, there is just one outcome in this event), so, assuming independence of the flips, the probability is $1/2^{10}$.

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The number of elements in your sample space is not right. Think if there is only one coin flip are there 1! ways of ordering them? You are right in that there is only one outcome with 5heads and tails in that order.

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Write down any (repeat any!) sequence of H and T for a total of 10 letters. Any sequence you write has the same probability; some just look prettier (to some) than others...

H H H H H T T T T T

H H H H H H H H H H

H H T H T T T H H T

All have exactly the same probability: $\frac{1}{2^{10}}$

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