13.06 The multiplication rule
Probability of independent events
Recall that two events are independent if the occurrence of one event does not affect the probability of the other occurring.
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 below.
If two events $A$A and $B$B are independent, then the probability of both events occuring is the product of the probability of A and the probability of B. In other words,
$P(A)\times P(B)=\text{P(A and B)}$P(A)×P(B)=P(A and B)
The converse is also true. This means you 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 $6$6-sided die is rolled. Find the probability of flipping a tail on the coin and rolling a $4$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 2
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 3
In a school, $25%$25% of students ride skateboards and $20%$20% of students wear glasses. One student is selected at random. What is the probability that the student: wears glasses and rides a skateboard? does not wear glasses and does not ride a skateboard? wears glasses and does not ride a skateboard? does not wear glasses and does ride a skateboard?
Question 4
An archer makes two attempts to hit a target, and the probability that he hits the target on any one attempt is $\frac{1}{3}$13. What is the probability that the archer misses the target? What is the probability that the archer will miss the target on both attempts? $\frac{8}{9}$89 $\frac{1}{9}$19 $\frac{2}{3}$23 $\frac{4}{9}$49
Probability of dependent events
Recall 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.
If two events $A$A and $B$B are dependent, then the probability of both events occurring is the product of the probability of $A$A and the probability of $B$B after $A$A occurs.
$P(A)\times\text{P(B following A)}=\text{P(A and B)}$P(A)×P(B following A)=P(A and B)
Worked examples
Question 5
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 6
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: $P(J,Q,K)$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 7
In a game of Blackjack, a player is dealt a hand of two cards from the same standard deck. What is the probability that the hand dealt: Is a Blackjack? (A Blackjack is an Ace paired with 10, Jack, Queen or King.) Has a value of 20? (10, Jack, Queen and King are all worth 10. An Ace is worth 1 or 11.)
Question 8
Vanessa has $12$12 songs in a playlist. $4$4 of the songs are her favorite. She selects shuffle and the songs start playing in random order. Shuffle ensures that each song is played once only until all songs in the playlist have been played. What is the probability that: the first song is one of her favorites? two of her favorite songs are the first to be played? at least one of her favorite songs is played in the first three?