0.4 0.8 0.6 0.2 .
a If Frank dives to the left on a penalty he defends in today’s game, what is the probability that he will dive to the right for the next penalty?
b Suppose there is a penalty shoot-out that consists of each team taking five penalty shots at goal.
The player taking the fifth penalty kicks to the right. If Frank had dived to the left on the first penalty, what is the probability that he dives to the right on the fifth penalty? Give answers accurate to 2 decimal places.
9 Assume that the probability of a particular football team winning its next game is 0.75 if it won its previous game and 0.55 if it lost its previous game.
a If the team was successful in the opening game of the season, calculate the probability that it will win:
i the second game of the season ii the fifth game of the season iii the tenth game of the season iv the twelfth game of the season.
Give answers accurate to 4 decimal places.
b Repeat the calculations to find the corresponding probabilities if the team lost its opening game.
c Can you predict, without further calculation, what is likely to happen in the last game of the season? (Assume the season lasts for twenty-four games.)
10 Tasha likes to vary her gym routines in a special way. If she does a cycling class one day, the probability that she will do a Pilates session the next day is 0.6, and if she does a Pilates session one day, the probability she does cycling next day is 0.75. Assume that Tasha goes to the gym every day and does only cycling or Pilates.
a Write down the transition matrix for this situation.
b If Tasha cycles on Saturday, what is the probability that she also cycles the next Tuesday?
c What is the probability that she does Pilates on that Tuesday?
d What is the probability that she does Pilates on the next Tuesday?
11h independent events
Two events A and B are independent if each event has no effect on the likelihood of the other.
Consider two independent events A and B, where event A follows event B. If the probability of event A is unaffected by event B, then we can say that the probability of A, given B has happened, is the same as the probability of A (whether or not B has happened), or, using symbols:
Pr(A | B) = Pr(A) [1]
But A B| = A B∩ Pr( ) Pr ( B )
Pr ( ) using the conditional probability formula. Rearranging the above equation we have:
Pr(A ∩ B) = Pr(B) Pr(A | B) [2]
Note: Equation [2] has wide application in probability. It may be extended and
interpreted as: ‘When calculating the probability of a chain of events, you may simply multiply by the probability of the next event, as long as the effect of previous events is taken into account’.
Substituting [1] into [2] we have:
Pr(A ∩ B) = Pr(A) Pr(B)
1. Pr(A ∩ B) means the probability of events A and B occurring.
2. If Pr(A ∩ B) = Pr(A) × Pr(B), then the events A and B are independent.
One of the 12 outcomes possible when a coin and a die are simultaneously tossed is a Head for the coin and a 5 on the die. The number 5 obtained with the die does not come about because the coin comes up a Head, and getting a Head with the coin is not a result of the number 5 appearing uppermost on the die.
We can verify the expression given for independent events by looking further at the example of the coin and die.
What is the probability of getting a Tail and a number from 3 to 4 inclusive?
The event space is ξ = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}, with n(ξ ) = 12.
Let A be the event ‘getting a Tail’ and B be the event ‘getting a number from 3 to 4 inclusive’.
Then A ∩ B = {T3, T4}, n(ξ) = 12, so Pr(A ∩ B) = n(A ∩ B)
n(ξ ) = 122 = 16. Now Pr(A) = 12 and Pr(B) = 26 = 13, so Pr(A) × Pr(B) = 12 × 13 = 16. So Pr(A ∩ B) = Pr(A) × Pr(B); thus the two events are independent.
WorkEd ExamplE 29
Given that events A and B are independent, find the value of x if Pr(A) = 0.55, Pr(B) = 0.6 and Pr(A ∩ B) = x.
Think WriTE
1 Write the formula for independent events. Pr(A ∩ B) = Pr(A) × Pr(B)
2 Substitute the given information. x = 0.55 × 0.6
3 Simplify. x = 0.33
WorkEd ExamplE 30
Show that if Pr(A) = 0.5, Pr(B) = 0.8 and Pr(A ∪ B) = 0.9, then A and B are independent.
Think WriTE
1 Use the Addition Law for probabilities to find
Pr(A ∩ B). Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)
0.9 = 0.5 + 0.8 − Pr(A ∩ B) Pr(A ∩ B) = 0.4
2 Calculate Pr(A) × Pr(B). Pr(A) × Pr(B) = 0.5 × 0.8
= 0.4
Since Pr(A ∩ B) = Pr(A) × Pr(B), A and B are independent events.
WorkEd ExamplE 31
Two spinners each labelled with the numbers 1, 2, 3 are spun.
A is the event ‘an odd number with the first spinner’.
B is the event ‘an even number with the second spinner’.
C is the event ‘an odd number from each spinner’.
a Calculate Pr(A), Pr(B) and Pr(C ).
b Decide which of the pairs of events AB, AC, BC is independent.
Think WriTE
a 1 List ξ, A, B and C. aξ = {11, 12, 13, 21, 22, 23, 31, 32, 33}
A = {11, 12, 13, 31, 32, 33}, B = {12, 22, 32}, C = {11, 13, 31, 33}
2 Calculate Pr(A), Pr(B) and Pr(C ). Pr(A) = 69 = 23, Pr(B) = 39 = 13, Pr(C) = 49
Spinner 2 Spinner 1
1 2
3
1 2
3
b Check to see if
Pr(A ∩ B) = Pr(A) × Pr(B), Pr(A ∩ C) = Pr(A) × Pr(C ), Pr(B ∩ C) = Pr(B) × Pr(C ).
bA ∩ B = {12, 32}, Pr(A ∩ B) = 29 Pr(A) × Pr(B) = 23× 13
= 29
Pr (A ∩ B) = Pr(A) × Pr(B), so A and B are independent.
A ∩ C = {11, 13, 31, 33}, Pr(A ∩ C ) = 49 Pr(A) × Pr(C ) = 23 × 49
= 278
Pr(A ∩ C ) ≠ Pr(A) × Pr(C ), so A and C are not independent.
B ∩ C = ϕ, Pr(B ∩ C ) = 0 Pr(B) × Pr(C ) = 13 × 49 = 274
Pr(B ∩ C ) ≠ Pr(B) × Pr(C ), so B and C are not independent.
When the probabilities of all possible outcomes are not equally likely, the probability of each outcome is placed on the corresponding branch of the tree diagram. When each branch is representing an outcome from independent events, you can follow the branches and multiply the probabilities together.
WorkEd ExamplE 32
A moneybox contains three $1 coins and two $2 coins. The moneybox is shaken; one coin falls out and is put back in the box. This is repeated twice more. If each coin has an equal probability of falling out:
a represent this information on a tree diagram b calculate the probability of getting three $1 coins c calculate the probability of getting at least two $2 coins.
Think WriTE/draW
a 1 There are three $1 coins and fi ve coins altogether. The probability of a $1 coin falling out is 35 = 0.6.
a
2 There are two $2 coins and fi ve coins altogether. The probability of a $2 coin falling out is 25 = 0.4.
3 Place the probability of each outcome on the
corresponding branch of the tree diagram. $1
$2
$1
$2
$2
$1
$1 $1, $1, $1
$2 $1, $1, $2
$1 $1, $2, $1
$2 $1, $2, $2
$1 $2, $1, $1
$2 $2, $1, $2
$1 $2, $2, $1
$2 $2, $2, $2 0.6
0.4
0.6 0.4 0.6 0.4 0.6 0.4 0.6 0.4 0.6
0.4 0.6 0.4
b Multiply the probabilities obtained from the tree
diagram corresponding to three $1 coins. bPr(three $1 coins) = 0.6 × 0.6 × 0.6
= 0.216 c 1 Outcomes corresponding to ‘at least two $2
coins’ are ($1, $2, $2), ($2, $1, $2), ($2, $2, $1) and ($2, $2, $2).
c
2 Calculate and add the probabilities. Pr(at least two $2 coins)
= Pr($1, $2, $2 or $2, $1, $2 or $2, $2, $1 or $2, $2, $2)
= (0.6 × 0.4 × 0.4) + (0.4 × 0.6 × 0.4) + (0.4 × 0.4 × 0.6) + (0.4 × 0.4 × 0.4)
= 0.096 + 0.096 + 0.096 + 0.064
= 0.352
WorkEd ExamplE 33
Christos estimates his chances of passing Maths, Science and English as 0.75, 0.6 and 0.5 respectively.
a Represent this information on a tree diagram.
b Assuming the events are independent, calculate the probability that:
i he passes all three subjects ii he passes at least Maths and English iii he passes at least one subject.
Think WriTE/draW
a 1 Name the three events. a Let M, S, E be the events ‘passing Maths’,
‘passing Science’ and ‘passing English’
respectively.
2 Calculate Pr(M ′), Pr(S ′) and Pr(E ′). M ′ is the event of failing Maths.
Pr(M) = 0.75 Pr(M′) = 0.25 Pr(S) = 0.6 Pr(S′) = 0.4 Pr(E ) = 0.5 Pr(E′) = 0.5
3 Use the information to draw the tree diagram.
S
S'
S
S' M'
M
E MSE E' MSE' E MS'E E' MS'E'
E M'SE E' M'SE' E M'S'E E' M'S'E' Maths Science English
0.75
0.25
0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.4
0.6 0.6
0.4
b i Multiply the probabilities corresponding
to passes in all three subjects. b iPr(MSE) = Pr(M) × Pr(S) × Pr(E)
= 0.75 × 0.6 × 0.5
= 0.225 ii 1 We require that Christos pass both
Maths and English and either pass or fail Science.
iiPr(MSE or MS ′E)
= Pr(MSE) + Pr(MS ′E)
2 These events are independent, so we may
multiply the individual probabilities. = Pr(M) × Pr(S) × Pr(E) + Pr(M) × Pr(S ′) × Pr(E)
= 0.75 × 0.6 × 0.5 + 0.75 × 0.4 × 0.5
3 Simplify. = 0.375
iii Passing at least one subject is the complement of
failing all three subjects. iii
Pr(passes at least one subject)
= 1 – Pr( M′S′E′)
= 1 – 0.25 × 0.4 × 0.5
= 1 – 0.05
= 0.95
TUTorial eles-1453
Worked example 33