hernandezlarry1234
Probability of getting 2 heads and 4 tails when flipping a coin 6 times
hernandezlarry1234,
I'm happy to help.
I saw your post without an answer so far, so I thought I would help.
Think about it this way. In the six flips, how many ways can we place the two heads? We need a combination for this.
See:
GMAT Permutations and CombinationsGMAT Math: Calculating CombinationsHere, the number of ways is 6C2 = 15. There are 15 cases.
In each case, we have six flips, and each one is either H or T. Both have a probability of 1/2, so any one case would have a probability of
\(\frac{1}{2^6} = \frac{1}{64}\)
Since there are 15 cases with identical probabilities, the total probability is
\(\frac{15}{2^6} = \frac{15}{64}\)
This particular setup is known as the Binomial situation in Probability. Here's a free lesson about it:
Binomial SituationDoes all this make sense?
Mike