The chain rule is used when you have multiple trials, meaning that you want to measure several events one after another. In these cases you need to multiply the probability of the first event by the probability of the second event. It’s important to pay attention to whether the events are independent of each other or not.

Rule

When $A$ and $B$ are independent events, you get

$$P\phantom{\rule{-0.17em}{0ex}}\left(A\cap B\right)=P\phantom{\rule{-0.17em}{0ex}}\left(A\right)\cdot P\phantom{\rule{-0.17em}{0ex}}\left(B\right)$$ |

Example 1

**In mathematics, births are considered to be independent events. $P\phantom{\rule{-0.17em}{0ex}}\left(\text{boy}\right)=\text{}0.514\text{}$ and $P\phantom{\rule{-0.17em}{0ex}}\left(\text{girl}\right)=\text{}0.486\text{}$. A family wants four children. What is the probability of the first two being boys and the next two being girls? **

In this case you don’t have to worry about the order of the births of each single baby, just that there are two boys first, and then two girls. The complete event is boy, boy, girl, girl. You can write it as $BBGG$. The calculation becomes

$$\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(BBGG\right)& =P\phantom{\rule{-0.17em}{0ex}}\left(B\right)\cdot P\phantom{\rule{-0.17em}{0ex}}\left(B\right)\cdot P\phantom{\rule{-0.17em}{0ex}}\left(G\right)\cdot P\phantom{\rule{-0.17em}{0ex}}\left(G\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =P\phantom{\rule{-0.17em}{0ex}}{\left(B\right)}^{2}\cdot P\phantom{\rule{-0.17em}{0ex}}{\left(G\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.51{4}^{2}\cdot 0.48{6}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.0624\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$The probability of having two boys and then two girls is $6.24$ %.

Rule

When the events $A$ and $B$ depend on each other, you get

$$P\phantom{\rule{-0.17em}{0ex}}\left(A\cap B\right)=P\phantom{\rule{-0.17em}{0ex}}\left(A\right)\cdot P\phantom{\rule{-0.17em}{0ex}}\left(B\mid A\right)$$ |

Example 2

**A bag of jelly beans has 4 red beans, 5 green ones and 7 yellow ones. You’re going to draw three jelly beans from the bag and eat them. What is the probability of drawing three yellow jelly beans? **

There’s $4+5+7=16$ jelly beans in total. After eating one yellow jelly bean, there is one yellow jelly bean less in the bag (6), and the total number of jelly beans has also been reduced by one (15). In this way the number of jelly beans in the bag will decrease each time you eat one. That makes the calculation

$$\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(\text{3yellow}\right)& =P\phantom{\rule{-0.17em}{0ex}}\left(\text{firstisyellow}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\cdot P\phantom{\rule{-0.17em}{0ex}}\left(\text{secondisyellow}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}\cdot P\phantom{\rule{-0.17em}{0ex}}\left(\text{thirdisyellow}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7}{16}\cdot \frac{6}{15}\cdot \frac{5}{14}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.0625\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

$$\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(\text{3yellowbeans}\right)& =P\phantom{\rule{-0.17em}{0ex}}\left(\text{firstisyellow}\right)\cdot P\phantom{\rule{-0.17em}{0ex}}\left(\text{secondisyellow}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\cdot P\phantom{\rule{-0.17em}{0ex}}\left(\text{thirdisyellow}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{7}{16}\cdot \frac{6}{15}\cdot \frac{5}{14}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.0625\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

The probability of eating three yellow jelly beans in a row, without grossing your friends out by putting them back in the bag after you’ve eaten them. is $6.25$ %. To find this, we had to calculate the probability of drawing a yellow jelly bean from the bag three times, each time with different numbers and answers. It’s pretty clear that it would be easier to find the answer if the events were independent of each other.