8.06 Probability of compound events

Remember that two events are independent if the occurrence of one event does not affect the probability of the other occurring.

For example, to find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, and then multiply the probabilities. This multiplication rule is written as:P(A \text{ and }B)= P(A) \times P(B)

Let's look at an example:

A coin is tossed and a single 6-sided die is rolled. Find the probability of flipping a tail on the coin and rolling a 4 on the die.

Let P(\text{tail}) be the probability of flipping a tail on the coin and P(4) be the probability of rolling a 4 on the die.

We know that the probability of flipping a tail is \dfrac{1}{2} so P(\text{tail})=\dfrac{1}{2}.

We also know that rolling a 4 on a die has a probability of \dfrac{1}{6} so let P(4)=\dfrac{1}{6}

We know that two events are dependent if the outcome of the first event affects the outcome of the second event in such a way that the probability is changed. All 'without replacement' events are dependent.

When we calculate the probabilities of dependent events we often have to adjust the second probability to reflect that the first event has already occurred.

Let's look at an example:

Three cards are chosen at random from a deck of 52 cards without replacement. What is the probability of choosing 3 kings?

The outcome of the first draw affects the outcome of the next draw. So they are dependent.

In a deck of of 52 cards, there are 4 kings. So on the first draw, the probability of drawing a king is \dfrac{4}{52}. If we don't replace the card, what is the probability of drawing a king on the second draw? Now we have a total of 51 cards, and only 3 kings left, so the probability is \dfrac{3}{51}, as shown in the image below.

We still multiply the probabilities of each event together, but as you can see each event has a different probability because they depend on the previous event occuring.

Examples

A tree diagram is useful in tracking two-step experiments. The important components of the tree diagram are:

• Branch

• Probability (used when the events have unequal probabilities)

• Outcome

When the outcomes are not equally likely the probability will be written on the branches.

The sum of the probabilities on branches from a single node should sum to 1 or 100\%.

When a single trial is carried out, we have just one column of branches.

Here are some examples. None of these have probabilities written on the branches because the outcomes are equally likely.

Here are some examples that have probabilities on the branches, because they do not have an equal chance of occurring.

When more than one experiment is carried out, we have two (or more) columns of branches.

This tree diagram shows the outcomes and sample space for tossing a coin twice. \text{T} is for Tails and \text{H} is for heads.

To calculate the probability of HH we would multiply \dfrac{1}{2} \times \dfrac {1}{2} to get \dfrac{1}{4}.

Here is an example that have probabilities on the branches. This probability tree diagram shows the outcomes of playing two games of tennis where the probability of winning is 0.3 and the probability of losing is 0.7.

The probabilities of the events are multiplied along each branch, for example the probability of winning both games is 9\% which is found by \\ 0.3 \times 0.3 = 0.09

To find the probability of at least 1 win, we could do:\text{P(win, lose)} + \text{P(lose, win)} + \text{P(win, win)} = 21\% + 21\% + 9\% = 51\%

When we have dependent events that are "without replacement" and depend on the event that happened previously, we need to take care when drawing the tree diagram.

For example:

The probability of drawing a red card from a standard pack of 52 cards is \dfrac{26}{52} = \dfrac{1}{2}. If we do draw a red card and choose to select a second card without replacement there are only 25 red cards left, but there are still 26 black cards. And there are only 51 cards left in the entire deck. The probability of selecting a second red card is \dfrac{25}{51}. This can be seen in the top branches of the tree diagram.

Examples

Example 2

On the island of Timbuktoo, the probability that a set of traffic lights shows red, yellow or green is equally likely. Christa is traveling down a road where there are two sets of traffic lights.

What is the probability that both sets of traffic lights will be red?

Worked Solution

Create a strategy

Find the number of favorable outcomes red and red or RR in the probability tree. From the tree diagram we know that each event is equally likely because the probabilities are not written on the branches.

Apply the idea

In the first event, there are 3 possible outcomes and only 1 is red. So the probability for getting red is \dfrac{1}{3}.

We can also see that for the second event the probability of getting red is the same, so another chance of \dfrac{1}{3}.

\displaystyle P(RR)	\displaystyle =	\displaystyle \dfrac{1}{3} \times \dfrac{1}{3}
\displaystyle P(RR)	\displaystyle =	\displaystyle \dfrac{1}{9}	multiply the probabilities

Example 3

Han owns four green ties and three blue ties. He selects one of the ties at random for himself and then another tie at random for his friend.

The color combinations of the two ties he selects are shown in this tree diagram.

a

Write the probabilities for the outcomes on the edges of the probability tree diagram.

Worked Solution

Create a strategy

The probability of picking a tie will be given by:\dfrac{\text{Number of colored ties}}{\text{Total number of ties}}

Apply the idea

Initially, Han has a choice of 4 green and 3 blue ties, out of a total of 7 ties.

So on the first set of branches: P(G)=\dfrac{4}{7} and P(B)=\dfrac{3}{7}.

Assuming his first selection was green, Han would then have a choice of 3 green and 3 blue ties, out of a total of 6 ties.

So on the top set of branches after G: P(G)=\dfrac{3}{6} and P(B)=\dfrac{3}{6}.

Assuming his first selection was blue, Han would then have a choice of 4 green and 2 blue ties, out of a total of 6 ties.

So on the bottom set of branches after B: P(G)=\dfrac{4}{6} and P(B)=\dfrac{2}{6}.

So our probability tree would be as follows:

b

What is the probability that Han selects a blue tie for himself?

Worked Solution

Create a strategy

Follow the path to the first B in the probability tree diagram.

Apply the idea

Using the tree diagram from part (a), the probability that Han selects a blue tie for himself is \dfrac{3}{7}.

c

Calculate the probability that Han selects two green ties.

Worked Solution

Create a strategy

To find the probabilities follow the top path to GG and multiply the probabilities.

Apply the idea

The probability that the first tie was green is \dfrac{4}{7}, and the probability that the second tie was also green is \dfrac{1}{2}.

\displaystyle P(GG)	\displaystyle =	\displaystyle \dfrac{4}{7} \times \dfrac{1}{2}	Multiply the probabilities
	\displaystyle =	\displaystyle \dfrac{4}{14}	Evaluate
	\displaystyle =	\displaystyle \dfrac{2}{7}	Simplify

The probability that Han selects two green ties is \dfrac{2}{7}.