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?
Best Answer
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.