0
$\begingroup$

If a fair coin is tossed three times what is the probability that heads shows once and tails shows twice?

I thought the answer would be $1/4 + 1/2$ because the probability of getting heads is $1/2$ and the probability of getting tails twice is $1/2 * 1/2 = 1/4,$ but this is incorrect. I am not sure what I did wrong.

$\endgroup$
1
  • $\begingroup$ More broadly, you only add probabilities if you're finding the probability of any of a set of mutually exclusive outcomes. That isn't the case here; you're not finding the probability of either getting one heads, or else of getting two tails. (And even in that case, the details of the exact situation can affect what you do with the individual probabilities.) $\endgroup$
    – Brian Tung
    Commented Jun 25, 2020 at 0:28

3 Answers 3

Reset to default
1
$\begingroup$

This can happen in $\dbinom{3}{1}$ ways. Therefore probability $= 3 \times \dfrac{1}{2} \times \dfrac{1}{2}\times \dfrac{1}{2}$

$\endgroup$
1
$\begingroup$

Each flip of a coin has a $\frac12$ chance to land on Heads and the same for Tails. Hence, if we flip it three times, the chance of getting any particular configuration (e.g. HTH in that order) is $\frac12 \cdot \frac12\cdot\frac12=\frac18$.

How many of the $8$ configurations you can possibly get have two tails and one heads?

$\endgroup$
1
$\begingroup$

The answers already present explain how to get it correct, but don't explain why it's wrong. I'll explain here.

There are two types of probability: dependent and independent.

  • Dependent probability is where a probability depends on another. For example, this problem uses dependent probability. The probability depends on all cases: 1 head and 2 tails. We use multiplication to calculate dependent probability.
  • Independent probability is where a probability is not dependent on another. For example, the probability that we get 1 head or less uses independent probability. The probability of 1 head and 0 heads are different, so we add the two probabilities. (Note that when calculating each individual probability, we use dependent probability.)

Now let's solve the problem with our knowledge. You are right that the probability of heads is $\frac{1}{2}$ and tails is $\frac{1}{2}\cdot \frac{1}{2} = \frac{1}{4}$. You used DP (dependent probability) right there to get the probability of two tails! However, since our final probability depends on both the heads and tails for one case, we multiply them instead of add.

Next, we have to multiply by $3$ because there are three ways to order one head and two tails: $HTT$, $THT$, and $TTH$. In math terms this is $\binom{3}{1}$, read as 3 choose 1. You can research this yourself later.

Therefore, the answer is $\boxed{\frac{3}{8}}$.

-FruDe

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .