9.04 Probability of compound events

Can these things happen at the same time?

Is it possible?

When we think of things happening, we know some things can't occur at the same time. You can't be sleeping at the same time as you are awake. If you are standing up, you are not sitting down.

Then there are things that may happen at the same time. You might be standing up, or you might be eating. It's also possible that you are standing up and eating! 

From the three arrows below, is it possible to choose an arrow that is straight and green?

While we can choose a green arrow and we can choose a straight arrow, we cannot choose one that satisfies both options. The only straight arrow is red. The only green arrow is curved. Therefore, none of the options satisfies both conditions we were looking for.

 

Only two outcomes

Sometimes there are only two possibilities that could occur. If you think of tossing a coin, we can only toss a head or a tail. If we only have one coin, there are no other options that could occur at the same time.

In our first video, we will look at events where there can only be two possibilities.

 

More than two outcomes

Sometimes, such as when you throw a die, you have more things that could happen. A die has $6$6 sides. If we only have one die, then we can only roll a $1,2,3,4,5$1,2,3,4,5 or $6$6. We can't roll a $12$12, as we only have one die.

A deck of cards has $52$52 cards. Within a deck of cards, there are different colors, numbers and patterns. This means there are many possibilities of things that could happen. In the second video, we look at having different options, to see what is possible.

 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3

When does one event affect another?

Independent or dependent?

When we look at the chance of something happening, there are ways in which the chance, or probability could be affected by what has already happened.

Independent events

An independent event means that the chances of that event happening are not changed by what happened before. For example, when we flip a coin, there is always a $\frac{1}{2}$12​ chance that it will land on heads. It doesn't matter whether you tossed a head or tail before. 

To see how this works, let's work through an example where you have four different colored lunchboxes and look at the chance of a particular color being chosen each day.

 

Dependent events

A dependent event means that the chance of that event changes depending on what happened before. For example, let's say you have an electronic claw machine game full of all different prizes. If you pull a toy car out, could someone else choose that same prize? 

Let's look at your lunchbox colors again and see how the chance of picking a particular color changes when you don't bring them home at the end of the day. 

 

Practice questions

Question 4

Question 5

Probability of independent events

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

For example,

• Tossing a coin is an independent event.  It really doesn't matter what the last 100 tosses have been, the next toss has a 50% chance of being a head and 50% chance of being a tail.  In fact any type of event that operates with replacement, or consecutive equally likely events is independent.
• Tossing a coin and then rolling a die are independent events, because they use completely different objects.  The die is not affected by the coin and vice versa.  Any type of events that use different objects are independent.  

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 defined symbolically as

$\text{P(A and B)}$P(A and B)$=$=$P(A)$P(A) x $P(B)$P(B), for independent events. 

This means you can also test for independence by verifying if P(A) x P(B)  = P(A and B).

 

Worked examples

Question 6

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.

Think:  $\text{P(Tail) }=\frac{1}{2}$P(Tail) =12​

$\text{P(4) }=\frac{1}{6}$P(4) =16​

Do:  

$\text{P(Tail and 4) }$P(Tail and 4)	$=$=	$\text{P(Tail) }\times\text{P(4) }$P(Tail) ×P(4)
	$=$=	$\frac{1}{2}\times\frac{1}{6}$12​×16​
	$=$=	$\frac{1}{12}$112​

 

Question 7

A nationwide survey found that $64%$64% of people in a small country town have unreliable internet access.  If $3$3 people are selected at random, what is the probability that all three have unreliable internet access?

Think:  The probability that all three have unreliable internet is the same as multiplying the probabilities P(unreliable internet)x P(unreliable internet)P(unreliable internet)

Do:  P(I) x P(I) x P(I) = $0.64\times0.64\times0.64=0.262144$0.64×0.64×0.64=0.262144 

Reflect:  Now why are the events independent here? You may think that in the case of selecting people from a population, we should not replace the first person before selecting the next. This would make the selections dependent. But consider the following:

Say the population is $1000000$1000000, and $640000$640000 of them do not have reliable internet access.

P(first person has unreliable internet access) = $\frac{640000}{1000000}=0.64$6400001000000​=0.64

If we remove one of these people from the population before the second draw, there would be $999999$999999 people left in the population, and $639999$639999 of them would have unreliable internet access:

P(second person has unreliable internet access) = $\frac{639999}{999999}=0.639$639999999999​=0.639..

Without going any further, you can see that the probability does not change that much.

So for a large sample space, the probability changes so little that we can consider successive events as being independent.

Practice questions

Question 8

Question 9

Probability of dependent events

We say that two events are dependent if the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed.

In calculating probabilities of dependent events we often have to adjust the second probability to consider the fact that the first event has already occurred.  All 'without replacement' events are dependent.

 

Worked examples

Question 10

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

Think:  On the first draw we have $52$52cards, and there are $4$4kings in the pack.  On the second draw, if we have already kept a king out - then we have $51$51 cards, and $3$3 kings still in the pack.  On the third draw, because we have already kept two kings out, then we have $50$50 cards and just $2$2 kings still in the pack.  

This results in the following probability of 3 kings being selected.

Do:  $P(3Kings)$P(3Kings) = $\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}$452​×351​×250​

= $\frac{1}{5525}$15525​

Reflect:  When you solve problems like this, you may find it helpful to draw a few boxes marking the $3$3 cards you are interested in, then writing the probabilities in each box. It's then easier to see what you need to multiply to calculate the answer.

Question 11

What is the probability of drawing (without replacement) a Jack, Queen and King:

a) in that order?

b) in any order?

Think:  

a)  A Jack first, well there are 4 possible Jacks out of 52 cards.

A queen next, there are 4 possible queens out of 51 cards (remember we kept out a card).

A king next, there are 4 possible kings out of 50 cards remaining.

Do:  P(J,Q,K) = $\frac{4}{52}\times\frac{4}{51}\times\frac{4}{50}=\frac{8}{16575}$452​×451​×450​=816575​

Think:  

b) For the first card I want either a J, Q or K (12 possible cards) from a total of 52.

For the second card, I will want one of the other values (8 possible cards) from a total of 51. For example, if the first card was a Jack, then I want any of the Kings or Queens.

For the third card I will want the last card I need (4 possible) to complete my set from a total of 50 cards.

Do:  P(J,Q, K in any order) = $\frac{12}{52}\times\frac{8}{51}\times\frac{4}{50}=\frac{16}{5525}$1252​×851​×450​=165525​

 

Practice questions

Question 12

Question 13