# Suppose that you toss a coin three times and roll three dice. What is the probability of obtaining two tails, one five, and at least one dice roll of less than three?

Feb 18, 2016

$\frac{1}{2} \cdot \frac{1}{12} = \frac{1}{24}$

#### Explanation:

So let's split this up into two parts: the coins and the dice. Since they are independent of each other, we can simply multiply the probabilities for each one at the end to get the total probability.

First, the coins. You flip a coin $3$ times, and you need to get two tails.
This can be split into two probabilities, the third flip is a head, and the third flip is a tail.
If it was a tail, you would have a $\frac{1}{2}$ probability to get each tail and (3!)/(3!) ways to order them. Thus,
1/2*1/2*1/2*(3!)/(3!) = 1/8
If it was a head, you would have a $\frac{1}{2}$ probability to get each tail and a $\frac{1}{2}$ probability to get the tail. You can order this in (3!)/(2!) ways. Thus,
1/2*1/2*1/2*(3!)/(2!) = 3/8
Added together, $\frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}$

So, there is a $\frac{1}{2}$ chance you will get two tails when you flip the coin $3$ times.

Now, let's consider the dice being rolled. You roll it three times. One must be a $5$, one must be less than $3$, and the third one can be anything. This will take a number of cases: the third one is a $5$, the third one is the same number as the second one, and the third one is a different number. Let's take each one.

If the third number is also a $5$. There is a $\frac{1}{6}$ chance to get a $5$ and a $\frac{2}{6} = \frac{1}{3}$ chance to get a number less than $3$. There are (3!)/(2!) ways to order the numbers. All multiplied,
1/6*1/6*1/3*(3!)/(2!) = 1/36

If the third number is the same number as the second number, there are two cases, but we can work around it. To get the $5$, the probability is $\frac{1}{6}$. To get any number less than $3$ (aka $1$ and $2$), the probability is $\frac{2}{6} = \frac{1}{3}$. To get that exact number (either the $1$ or the $2$) is $\frac{1}{6}$. There are (3!)/(2!) ways to order the numbers. All multiplied,
1/6*1/3*1/6*(3!)/(2!) = 1/36

If the third number is different then both numbers, then it must be a $3$, $4$, or $6$. The probability of the $5$ is $\frac{1}{6}$ and the chance to get a number less than $3$ is $\frac{2}{6} = \frac{1}{3}$. The probability of the third number to be one of the three numbers listed is $\frac{3}{6} = \frac{1}{2}$. Multiplied,
$\frac{1}{6} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{36}$
The three cases added together get you $\frac{1}{36} + \frac{1}{36} + \frac{1}{36} = \frac{1}{12}$

This multiplied by our probability of the coins gets $\frac{1}{12} \cdot \frac{1}{2} = \frac{1}{24}$