Probability

United’s manager estimates that the team has a 65% chance of winning any particular game and an 85% chance of not drawing any particular game.

a.  What are the manager’s estimates most likely to be based on?

b.  If the team plays 40 games this season, find the manager’s expectation of the number of games the team will lose.

c.  If the team loses one game more than the manager expects this season, explain why this does not necessarily mean that they performed below expectation.

a. From the team’s previous result

b `P_(lose) = 1 - [P_(w i n) + P_(draw)]`

`P_(lose) = 1 - [0.65 + (1 - 0.85)] = 0.2`

Expectation = `40*P_(lose) = 40 * 0.2`

Expectation = 8 games 

c. They may win some of the games that they are expected to draw

A numbered wheel is divided into eight sectors of equal size, as shown. The wheel is spun until it stops with the arrow pointing at one of the numbers. Axel decides to spin the wheel 400 times.

a.  Find the number of times the arrow is not expected to point at a 4.

b. How many more times must Axel spin the wheel so that the expected number of times that the arrow points at a 4 is at least 160?

a. Expectation = `n*`P(not 4) = `400*6/8 = 300`

b. `(400 + x) * 2/8 ge 240,`so he must spin the wheel at least 240 more times.

In a group of 25 boys, nine are members of the chess club (C ), eight are members of the debating club (D) and 10 are members of neither of these clubs. This information is shown in the Venn diagram.

a.  Find the values of a, b and c.

b.  Find the probability that a randomly selected boy is:

                  i.  a member of the chess club or the debating club

                  ii.  a member of exactly one of these clubs.

a. Solve 

`a + b + c + 10 = 25a + b = 9b + c = 8`

`a = 7b = 2c = 6`

b. i. The events are not mutually exclusive: 

`9/25 + 8/25 - 2/25 = 3/5`or `frac(7 + 2 + 6)(25) = 3/5`

ii. `frac(7 +6)(25) = 13/25`

Forty girls were asked to name the capital of Cuba and of Hungary; 19 knew the capital of Cuba, 20 knew the capital of Hungary and seven knew both.

a.  Draw a Venn diagram showing the number of girls who knew each of these capitals.

b.  Find the probability that a randomly selected girl knew:

i. the capital of Cuba but not of Hungary

ii. just one of these capitals.

a. 

b. i. `12/40 = 3/10`

ii. `frac(12 + 13)(40)= 5/8`

A garage repaired 132 vehicles last month. The number of vehicles that required electrical (E), mechanical (M) and bodywork (B) repairs are given in the diagram opposite. 

Find the probability that a randomly selected vehicle required:

a.  mechanical or bodywork repairs

b.  no bodywork repairs

c.  exactly two types of repair.

a. `frac(13 + 47 + 5 + 18 + 11 + 26)(132) = 10/11` or `frac(132-12)(132) = 10/11`

b. `frac(12 + 13 + 47)(312) = 6/11` or `frac(132 - (11 + 5 + 18 + 26)(132) = 6/11`

c. `frac(13 + 11 + 18)(132) = 7/22` or `frac(132-(12+47 + 26 + 5))(132) = 7/22`

Events X  and Y are such that P(X) = 0.5, P(Y) = 0.6 and P(`X cap Y)` = 0.2

a.  State, giving a reason, whether events X and Y are mutually exclusive.

b.  Using a Venn diagram, or otherwise, find `P(X capY)`.

c.  Find the probability that X or Y, but not both, occurs.

a. X and Y are not mutually exclusive because `P(X cap Y) ne 0`

Also, X and Y are not mutually exclusive because `P(X cap Y) ne 0` and because `P(X cap Y) ne P(X) + P(Y)`

b. 0.5 + 0.6 - 0.2 = 0.9 

c. (0.5 - 0.2) + (0.6 - 0.2) = 0.7 

Each of 27 tourists was asked which of the countries Angola (A), Burundi (B) and Cameroon (C) they had visited. Of the group, 15 had visited Angola; 8 had visited Burundi; 12 had visited Cameroon; 2 had visited all three countries; and 21 had visited only one. Of those who had visited Angola, 4 had visited only one other country. Of those who had not visited Angola, 5 had visited Burundi only. All of the tourists had visited at least one of these countries.

a.  Draw a fully labelled Venn diagram to illustrate this information.

b. Find the number of tourists in set B′ and describe them.

c.  Describe the tourists in set `(A cup B) cap C'`and state how many there are.

d.  Find the probability that a randomly selected tourist from this group had visited at least two of these three countries.

a. 

b. 9 + 3 + 7 = 19, they had not visited Burundi

c. They had visited Angola or Burundi, but not Cameroon;

9 + 1 + 5 = 15

d. `frac(1 +2 +3 + 0)(27) = 2/9` or `frac(27 - (9+5+7))(27) = 2/9`

The probabilities that a team wins, draws or loses any particular game are 0.6, 0.1 and 0.3, respectively.

a . Find the probability that the team wins at least one of its next two games.

b. If 2 points are awarded for a win, 1 point for a draw and 0 points for a loss, find the probability that the team scores a total of more than 1 point in its next two games.

a. `P(W, W') + P(W', W) + P(W, W) = 0.6*0.4 + 0.4*0.6 + 0.6* 0.6 = 0.84`

“At least one” has the same meaning as “not none”, so we could use `1- P(W', W') = 1 - 0.4*0.4 = 0.84`

b. `1- [P(L,L) + P(L, D) + P(D,L)] = 1- [(0.3*0.3) + (0.3*0.3) + 0.1 *0.3)] = 0.85`

A biased die in the shape of a pyramid has five faces marked 1, 2, 3, 4 and 5. The possible scores are 1, 2, 3, 4 and 5 and `P(x) = frac(k-x)(25)`, 

where k  is a constant.

a. Find, in terms of k, the probability of scoring:

i. 5 

ii. less than 3 

b. The die is rolled three times and the scores are added together. Evaluate k and find the probability that the sum of the three scores is less than 5.

a. i. Substituting x = 5 gives 

`P(5) = frac(k-5)(25)`

ii. P(score < 3) = P(1 or 2) = `frac(k-2)(25) + frac(k-1)(25) = frac(2k-3)(25)`

b. Sum of probabilities is 

`frac(k -1 + k -2 + k -3 + k-4+ k-5)(25) = 1`

Giving 5k - 15 = 25, so k = 8 

P(sum <5) = P(1,1,1) + P(1,1,2) P(1,2,1) + P(2,1,1) 

P(sum <5) = `(7/25)^3n+ 3*(7/25*7/25*6/25) = 49/625`or 0.0784 

A,B and C are independent events, and it is given that `P(A capB) = 0. 35, P(B cap C) = 0.56` and `P(A capC) = 0.4`

a. Express  in terms of:

i. P(B) 

ii. P(C) 

b.  Use your answers to part a to find:

i. P(B) 

ii. P(A')

iii. `P(B' capC')`  

a.i. `P(A) = 0.35/ (P(B))`

ii. `P(A) = 0.4/(P(C)`

b.i. `P(B' capC')`  = `P(B)*P(C) = 0.56`, so P(B) = `0.56/(P(C)`

`P(C) = frac(0.4*P(B))(0.35` gives `P(B) = frac(0.56*0.35)(0.4*P(B)`

`[P(B)]^2 = frac(0.56*0.35)(0.4) = 0.49`, so P(B) = 0.7

ii. P(A') = 1 - 0.5 = 0.5 

iii. `P(B' capC')` = `P(B')*P(C')`

`P(B' capC')`  = `(1-0.7)*(1-0.8) = 0.06`

Two ordinary fair dice are rolled.

Event X is the product of the two numbers obtained is odd.

EventY is the sum of the two numbers obtained is a multiple of 3.

a.  Determine, giving reasons for your answer, whether X and Y are independent.

b.  Are events X and Y mutually exclusive? Justify your answer.

a. P(X and Y) `1/12`, P(X) = `1/4`, P(Y) = `1/3` and `1/12 = 1/4 * 1/3`

P`(XcapY) = P(X)*P(Y)`, so X and Y are independent

b. X  and Y  are not mutually exclusive; X  and Y  both occur when, for example, 1 and 5 are rolled, i.e.

`P(XcapY) ne 0`

One hundred children were each asked whether they have brothers (B) and whether they have sisters (S). Their responses are given in the Venn diagram. 

Find the probability that a randomly selected child has:

a.  sisters, given that they have brothers

b.  brothers, given that they do not have sisters

c.  sisters or brothers, given that they do not have both.

a. 16 + 48 = 64 have brothers and 48 of these have sisters:

`48/64 = 3/4`

b. 16 + 12 = 28 do not have sisters, and 16 of these have brothers:

`16/28 = 4/7`

c. 16 + 24 + 12 = 52 do not have both, and 16 + 24 = 40 of these have brothers or sisters: 

`40/52 = 10/13`

Anya calls Zara once each evening before she goes to bed. She calls Zara’s mobile phone with probability 0.8 or her landline. The probability that Zara answers her mobile phone is 0.74, and the probability that she answers her landline is y. This information is displayed in the tree diagram shown.

a. Given that Zara answers 68% of Anya’s calls, find the value of y.

b. Given that Anya’s call is not answered, find the probability that it is made to Zara’s landline.

a. `(0.8*0.74) + (1-0.8)y = 0.68`

0.592 + 0.2y = 0.68, which gives y = 0.44 

b. P(call not answered) = (0.8*0.26) + (0.2*0.56) 

P(not answered) = `frac(0.2*0.56)(0.8*0.26+0.2*0.56)= 0.35`or `7/20`

Every Friday, Arif offers to take his sons to the beach or to the park. The sons refuse an offer to the beach with probability 0.65 and accept an offer to the park with probability 0.85. The probability that Arif offers to take them to the beach is x. This information is shown in the tree diagram

a. Find the value of x, given that 33% of Arif’s offers are refused

b. Given that Arif’s offer is accepted, find the probability that he offers to take his sons to the park

a. 0.65x + 0.15(1-x) = 0.33 gives x = 0.36 

b. P(offer accepted) = 0.36*0.35 + 0.64*0.85 

P(accepted) = `frac(0.64*0.85)(0.36*0.35 + 0.64*0.85) = 0.812` or `272/335`

Three friends, Rick, Brenda and Ali, go to a football match but forget to say which entrance to the ground they will meet at. There are four entrances, A, B, C and D. Each friend chooses an entrance independently.

• The probability that Rick chooses entrance A is `1/3`. The probabilities that he chooses entrances B,C or D are all equal.
• Brenda is equally likely to choose any of the four entrances.
• The probability that Ali chooses entrance C is `2/7` and the probability that he chooses entrance D is `3/5`. The probabilities that he chooses the other two entrances are equal.
  

  i. Find the probability that at least 2 friends will choose entrance B.

  ii. Find the probability that the three friends will all choose the same entrance.

i. 

P(at least two choose B) = P(two choose B) + P(thre choose B)

= P(abr') + P(ab'r) + P(a'br) + P(abr)

= `(2/35*1/4*7/9)+(2/35*3/4*2/9)+(33/35*1/4*2/9)+(2/35*1/4*2/9)` = `8/105` or 0.0762

ii. P(All A) + P(all B) +P(all C) + P(all D) 

`(2/35*1/4*3/9)+ (2/35*1/4*2/9)+(10/35*1/4*2/9)+(21/35*1/4*2/9) = 2/35`

Maria chooses toast for her breakfast with probability 0.85. If she does not choose toast then she has a bread roll. If she chooses toast then the probability that she will have jam on it is 0.8. If she has a bread roll then the probability that she will have jam on it is 0.4.

i.  Draw a fully labelled tree diagram to show this information.

ii.  Given that Maria did not have jam for breakfast, find the probability that she had toast.

i. Let T, B and X represent toast, bread and not jam, respectively:

ii. P(X) = P(T and X) + P(B and X) = `(0.85*0.2) + (0.15*0.6)`

P(X) = `frac(P(X and T))(P(X)) = frac(0.85*0.2)((0.85*0.2) +(0.15*0.6)) = 17/26`

Ronnie obtained data about the gross domestic product (GDP) and the birth rate for 170 countries. He classified each GDP and each birth rate as either ‘low’, ‘medium’ or ‘high’. The table shows the number of countries in each category.

One of these countries is chosen at random.

i. Find the probability that the country chosen has a medium GDP

ii.  Find the probability that the country chosen has a low birth rate, given that it does not have a
medium GDP

iii.  State with a reason whether or not the events ‘the country chosen has a high GDP’ and ‘the country chosen has a high birth rate’ are exclusive

One country is chosen at random from those countries which have a medium GDP and then a different country is chosen at random from those which have a medium birth rate.

iv. Find the probability that both countries chosen have a medium GDP and a medium birth rate

i. `frac(20 + 42 + 12)(170) = 37/85` or 0.435 

ii. `frac(3+35)(3+5+45+35+8) = 19/48`or 0.396

iii. They are (mutually) exclusive; 

P(high GDP and high birth rate) = 0 

iv. 

`42/(20+42 +12) * frac(41)(5+41+8)= 42/74*41/54= 287/666` or 0.431 

Two ordinary fair dice are rolled. Event A is ‘the sum of the numbers rolled is 2, 3 or 4’. Event B is ‘the absolute difference between the numbers rolled is 2, 3 or 4’.

a.  When the dice are rolled, they show a 1 and a 3. Explain why this result shows that events A and B are not mutually exclusive.

b.  Explain how you know that `P(A capB) = 1/18`

c. Determine whether events A and B are independent. 

a. Rolling a 1 and a 3 is favorable to A and to B

b. Exactly two of the 36 equally-likely outcomes i.e. (1,3) and (3,1), are favorable to A and to B, so

`P(A cap B) = 2/36 = 1/18`

c. P(A) = `1/6`, P(B) = `1/2`, `P(A capB) = 1/18`

`1/18 ne 1/6*1/2 = 1/12`

So `P(A capB) ne P(A)*P(B)`and, therefore, A and B arenot independent. 

Two ordinary fair dice are rolled. Event A is ‘the sum of the numbers rolled is 2, 3 or 4’. Event B is ‘the absolute difference between the numbers rolled is 2, 3 or 4’.

a.  When the dice are rolled, they show a 1 and a 3. Explain why this result shows that events A and B are not mutually exclusive.

b.  Explain how you know that `P(A capB) = 1/18`

c. Determine whether events A and B are independent. 

a. Rolling a 1 and a 3 is favorable to A and to B

b. Exactly two of the 36 equally-likely outcomes i.e. (1,3) and (3,1), are favorable to A and to B, so

`P(A cap B) = 2/36 = 1/18`

c. P(A) = `1/6`, P(B) = `1/2`, `P(A capB) = 1/18`

`1/18 ne 1/6*1/2 = 1/12`

So `P(A capB) ne P(A)*P(B)`and, therefore, A and B arenot independent. 

Bookings made at a hotel include a room plus any meal combination of breakfast (B), lunch (L) and supper (S). The Venn diagram opposite shows the number of each type of booking made by 71 guests on Friday.

a. A guest who has not booked all three meals is selected at random.

Find the probability that this guest:

i.  has booked breakfast or supper

ii.  has not booked supper, given that they have booked lunch.

b. Find the probability that two randomly selected guests have both booked lunch, given that they have both booked at least two meals.

a.i. `frac(7+24+6+16+5)(71-3) = 29/34`

ii. `frac(6+9)(6+9+5) = 3/4`

b. `frac(6+5+3)(7+6+5+3)*frac(6+5+3-1)(7+6+5+3-1) = 14/21*13/20= 13/20`

The following table shows the numbers of IGCSE (I) and A Level (A) examinations passed by a group of university students.

a.  For a student selected at random, find:

i. P(A < 4)

ii. P(I + A > 10)

b. Six students who all have at least three A Level passes are selected at random.

Find the greatest possible range of the total number of IGCSE passes that they could have.

a.i. `frac(6+2)(20+17+14+10+3)= 1/8`

ii. `frac(2+1)(1+6+2+1+1+1) = 1/4`

b. Minimum possible total is `6*5 = 30`

Maximum possible total is `10+9+9+8+8+8 = 52`

Greatest possible range is 52 - 30 = 22