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A coin is tossed six times. What is the probability of getting at least four heads on the tosses?

I have solved the problem like this:

probability of getting 2 tail = ${}_6C_4 \times (\frac{1}{2}^6)$

probability of getting 1 tail = $_6C_5 \times (\frac{1}{2}^6)$

There for the probability of getting least four heads on the tosses is the total probability minus the probability of getting the probability of getting 2 tail and probability of getting 1 tail.

Where am I wrong? Can you fix my probelm?

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    $\begingroup$ You should not write $(\frac12^6)$ if you mean $(\frac12)^6$. $\endgroup$ – Michael Hardy Aug 7 '14 at 16:55
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You are missing the probability that you get no tails--that is to say, the probability of getting at least four heads includes the event that you get all heads in $6$ tosses.

Also, implied in your question is the property that the coin is actually fair. Without this assumption, you cannot obtain a numerical result.

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I would like to add, that in that specific case, there is easier solution.

you could just calculate the probability of getting 3 tails and 3 heads

which is:

${}_6C_3 \times (\frac{1}{2})^6$.

if it not that case there are only 2 other cases:

  1. at least 4 heads.
  2. at least 4 tails.

because the probability of the above to cases is equal.than the answer is:

$(1 - {}_6C_3 \times (\frac{1}{2})^6) / 2$

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