# Independent coin flips and the binomial distribution

Four fair coins are flipped. If the outcomes are assumed independent, what is the probability that two heads and two tails are obtained.

The probability of success for either heads or tails is the same which is of course $$\frac{1}{2}$$.

I'm unsure of what to go from here, however, I know it has to incorporate the binomial distribution somehow. I also know, the general rule of thumb when you see the word "and" in a probability question, you multiply them.

Any help would be appreciated!

• "I also know, the general rule of thumb when you see the word 'and' in a probability question, you multiply them." Hopefully I can stop you right now and prevent you from making a mistake in the future. $Pr(A\cap B)$ is not equal to $Pr(A)\times Pr(B)$ in most scenarios. It is only equal if $A$ and $B$ are independent events. More generally $Pr(A\cap B) = Pr(A)\cdot Pr(B\mid A)$ – JMoravitz Sep 24 '18 at 2:07
• You should then have also heard a general rule of thumb to add if you saw the word "or." That is also incorrect. $Pr(A\cup B)$ is not equal to $Pr(A)+Pr(B)$ in most scenarios. It is only equal if $A\cap B$ is an (almost) impossible event (an event which occurs with probability zero). The more general true statement is $Pr(A\cup B)= Pr(A)+Pr(B)-Pr(A\cap B)$. – JMoravitz Sep 24 '18 at 2:08
• That being said, if you just want the result the distribution to be used here is called the binomial distribution. If you want to figure out the distribution on your own, you will break apart based on events detailing what the result of each individual flip are. Two heads and two tails were obtained occurs in one of the following ways: $(H,H,T,T), (H,T,H,T),(H,T,T,H),\dots,(T,T,H,H)$. Noting that these are mutually exclusive outcomes, we are allowed to simply add, and for each outcome we can simply multiply. – JMoravitz Sep 24 '18 at 2:12
• Possible duplicate of Probability of 3 Heads in 10 Coin Flips – JMoravitz Sep 24 '18 at 2:15