topic badge
Hong Kong
Stage 4 - Stage 5

Identifying Independent and Dependent Events


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$100 tosses have been, the next toss has a $50%$50% chance of being a head and $50%$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$\cap$$B)$B)$=$=$P(A)$P(A)$\times$×$P(B)$P(B), for independent events. The symbol that looks like a shoehorn is used in the case where we have one event AND another occurring.

This means we can also test for independence by verifying if $P(A)\times P(B)=\text{P(A and B)}$P(A)×P(B)=P(A and B).


Worked examples

Question 1

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

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

$\text{P(4) }=\frac{1}{6}$P(4) =16

$\text{P(Tail and 4) }$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



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?

$P(I)\times P(I)\times P(I)$P(I)×P(I)×P(I)$=$=$0.64\times0.64\times0.64=0.262144$0.64×0.64×0.64=0.262144 

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.


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 1

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

On the first draw we have $52$52 cards, and there are $4$4 kings 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$3 kings being selected:

$\text{P(3 Kings) }$P(3 Kings) $=$= $\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}$452×351×250
  $=$= $\frac{1}{5525}$15525

When we solve problems like this, it can be helpful to draw a few boxes marking the cards we are interested in, then writing the probabilities in each box. It is then easier to see what we need to multiply to calculate the answer.

Question 2

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

a) in that order?

b) in any order?

a)  A Jack first. There are $4$4 possible Jacks out of $52$52 cards.

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

A king next. There are $4$4 possible kings out of $50$50 cards remaining.



b) For the first card we want either a Jack, Queen or King ($12$12 possible cards) from a total of $52$52.

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

For the third card, we then want the last card needed ($4$4 possible) to complete the set from a total of $50$50 cards.

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 1

A fair coin is tossed 600 times.

  1. If it lands on heads 293 times, what is the probability that the next coin toss will land on heads?

  2. If it lands on heads once, what is the probability that the next coin toss will land on heads?

  3. Is the outcome of the next coin toss independent of or dependent on the outcomes of previous coin tosses?





Question 2

Vanessa has $55$55 songs in a playlist. She starts to play them on shuffle (this means each song will be played once until all songs have been played).

  1. What is the probability of Song A coming on first?


  2. What is the probability of Song A coming on last?


  3. Is the probability of a particular song playing dependent or independent of the position it plays in?






What is Mathspace

About Mathspace