Probability

I am having my exam soon so I'm doing a lot of questions. And I found difficult in the questions below. I hope someone can teach me how to do them. Thanks.

3)Two chess player, K1 and K2, are playing each other in a series of
games. The probability that K1 wins the first game is 0.3 . If K1 wins
any game, the probability that he wins the next is 0.4; otherwise the
probability is 0.2 . (I have cleared the first few parts so I do not
write those questions)
>>The result of any game can be a win for K1, a win for K2, or a draw.
The probability that any one game is drawn is0.5 , independent of the
result of all previous games. Find the probability that, after two
games,
(e)K1 won the first and K2 the second
(f)each won one game
(g)each has won the same number of games

4)Given that P(A)=0.75, P(B|A)=0.8 AND P(B|A')=0.6, calculate P(B) and
P(A|B).

7)Two events A and B are such that P(A)= 3/4, P(B|A)=1/5 and P(B'|A')
=4/7, By use of a tree diagram, or otherwise, find
a)P(A and B)
b)P(B)
c)P(A|B)

8)The probability of event A occuring is P(A)=13/25.The probability of
event B occuring is P(B)=9/25. The conditional probability of A
occuring given that B has occured is P(A|B)=5/9.
a)Determine the folllowing probabilities
iv)P(A'|B')
b)Determine P(A occurs or B does not occur) showing your working.

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Reply 1

Dilute

Holy.
Did you just copy and paste your whole practice paper on here?
At least try some of them!

Reply 2

authecroix123

zhang

Holy.
Did you just copy and paste your whole practice paper on here?
At least try some of them!

No, I didn't. As I have said, I have been doing a lot of questions as my exam is coming soon. These are all the questions that I cannot answer in the whole chapter. If you do not believe, you can check the A-level statistics book which endorsed by CIE. Those questions are from Exercises 4A, 4B and Miscellaneous exercise 4. There's just one more week left so I hope someone can help me! Thanks

Reply 3

Dilute

Okay.

1)Two fair dice are thrown simultaneously. Find the probability that
the two scores are the same.

= P(1,1) + P(2,2) + P(3,3) + P(4,4) + P(5,5) + P(6,6) = 6 x 1/36 = 6/36 = 1/6

14)A fair cubical dice is thrown four times. Find the probability that
a)all four scores are 4 or more,
=3xP(4 or 5 or 6)
= 3/6 x 3/6 x 3/6 = 1/8

b)at least one score is less than 4,
<4 = 1,2,3
4+ = 4,5,6
= P(<4, 4+, 4+) + P(4+, <4, 4+) + P(4+, 4+, <4) + P(<4, <4, 4+) + P(<4, 4+, <4) + P(4+, <4, <4) + P(<4, <4, <4)
= 7 x (3/6 x 3/6 x 3/6) = 7 x 1/8 = 7/8

c)at least one of the scores is a 6.
6 = 6
0 = not 6
= P(6, 0, 0) + P(0, 6, 0) + P(0, 0, 6) + P(6, 6, 0) + P(6, 0, 6) + P(0, 6, 6) + P(6, 6, 6)
= 3x(1/6 x 5/6 x 5/6) + 3x(1/6 x 1/6 x 5/6) + (1/6 x 1/6 x 1/6)
= 75/216 + 15/216 + 1/216
= 91/216

Reply 4

authecroix123

Thanks for helping me but your answers for the question 14 are wrong.
The answer should be
a)1/16
b)15/16
c)671/1296

I have figured it out.
The first part would be like this
P(all four scores are 4 or more) = 3/6 x 3/6 x 3/6 x 3/6
= 1/16
P(a least one score is less than 4) = 15 x (3/6 x 3/6 x 3/6 x 3/6)
= 15/16
I should not be having problem on the third part now. Thank you for helping me in understanding it.

Reply 5

Dilute

authecroix123

Thanks for helping me but your answers for the question 14 are wrong.
The answer should be
a)1/16
b)15/16
c)671/1296

Sorry!
I read the question wrong, I thought it was only thrown 3 times, not 4.

Reply 6

authecroix123

Thanks Zhang once again. I now know how to answer question 2 as well. I will delete any question(s) that has been answered. I hope other questions can be answered soon too! Thanks!

Reply 7

Dilute

6)Half of the A-level students in a community college study science
and 30% study mathematics. Of those who study science, 40% study
mathematics.
a)What proportion of the A-level students study both mathematics and
science?
40% of scientists do matsh = 40% of 50% of total students = 4/10 x 5/50 = 2/10 = 20%

b)Calculate the proportion of those students who study mathematics but
do not study science.
20% do maths and science. 30% do maths. 30-20 = 10%

Reply 8

authecroix123

zhang

I do not undertand part a. Why 4/10 x 5/50? What does the 5 refer to? Sorry for my ignorance, and thank you.

Reply 9

Jonny W

11)In a certain part of the world there are more wet days than dry
days. If a given day is wet, the probability that the following day
will also be wet is 0.8 . If a given day is dry, the probability that
the following day will also be dry is 0.6 .
(I have answered the first few parts so I just write the one I do not
know how to answer)
In one season there were 44 cricket matches, each played over three
consecutive days, in which the first and third days were dry. For how
many of these matches would you expect that the second day was wet?

Consider a randomly chosen match whose first day is dry. Define two events:

A = {second day is wet}
B = {third day is dry}

Then

P(A and B)
= 0.4*0.2
= 0.08

P(B)
= 0.4*0.2 + 0.6*0.6
= 0.44

P(A | B)
= P(A and B)/P(B)
= 0.08/0.44
= 2/11

So we would expect 44*(2/11) = 8 of the 44 matches to have a wet second day.

Reply 10

Jack

Here are my answers, post back if you don't understand them (Couldn't be bothered to do them all so I did the interesting ones only). I advise you to re-read the section explaining how to do these, especially look at page 71 with tree diagram.

Question 9:
iii) P(C or D or both occur) = P(A) + P(B) - P(A and B)
P(A and B) = P(A) * (B|A) or P(B) * (A|B)
5p + 4p - (1/5 * 5p) = 8p = 1/5
Therefore P = 0.025 = 1/40

Question Eleven:
P(DWD)/((P(DWD)+P(DDD) =2/11 2/11 * 44 = 8

Question 12:
Define the events, P(A) = Lesserprod comp or not (Bestbits)
P(B) = Fail, or not fail
So 0.01 = (N*1/50))+((1-N)*1/200)
0.01 = 1/50N + 1/200 - 1/200N
0.01 - 1/200 = 4/200N - 1/200N
Therefore N = (0.01 - 1/200) / (3/200)=1/3
And this is the prop for Lesser so 1 - N=2/3 = Bestbits

Question 15, Binomial Distribution:
B≈(24,1/36)
P(X>_1)
Therefore, using a binomial distribution table the probability is:
P(X>_1) = 0.4914038761

Reply 11

Dilute

authecroix123

I do not undertand part a. Why 4/10 x 5/50? What does the 5 refer to? Sorry for my ignorance, and thank you.

The Science students are 50% of the population. And 40% of those science students do maths too. Therefore, the amount that do both maths and science is 40% of 50% of the total population. 40% = 4/10 and 50% = 5/10
And 40% of 50% = 4/10 x 5/10

Reply 12

Jonny W

13)A game is played using a regular 12-faced fair dice, with faces
labelled 1 to 12, a coin and a simple board with nine squares as shown
in the diagram.
L _ _ _ * _ _ _ R
Initially, the coin is placed on "*". The game consists of rolling the
dice and then moving the coin one rectangle towards L or R according
to the outcome on the dice. If the outcome is a prime number(2,3,5,7,
or 11) the move is towards R, otherwise it is towards L. The game stops
when the coin reaches either L or R. Find, giving your answers correct
to 3 decimal places, the probability that the game
a)ends on the fourth move at R,
b)ends on the fourth move,
c)ends on the fifth move,
d)takes more than six moves.

(a)
(5/12)^4 = 0.030

(b)
(5/12)^4 + (7/12)^4 = 0.146

(c)
Label the squares as -4, -3, -2, -1, 0, 1, 2, 3, 4 (so -4 = L, 0 = * and 4 = R). The coin starts on an even number, then moves to an odd number, then to an even number, and so on until the game ends. After four moves, the game has ended or the coin is on -2, 0 or 2. So the game can't end on the fifth move. Probability = 0.

(d)
P(coin is at -2 after four moves) = 4C1 (7/12)^3 (5/12)

If the coin is at -2 after four moves, there is probability (7/12)^2 that the game will end on the sixth move.

P(coin is at 2 after four moves) = 4C1 (5/12)^3 (7/12)

If the coin is at 2 after four moves, there is probability (5/12)^2 that the game will end on the sixth move.

If the coin is at 0 after four moves, the game can't end on the sixth move.

--

P(game ends on the sixth move)
= [4C1 (7/12)^3 (5/12)] (7/12)^2 + [4C1 (5/12)^3 (7/12)] (5/12)^2
= 4 (7/12) (5/12) [(7/12)^4 + (5/12)^4]
= 0.142

P(game takes more than six moves to end)
= 1 - P(game ends on the fourth move) - P(game ends on the sixth move)
= 1 - 0.146 - 0.141
= 0.712

Reply 13

authecroix123

Jonny W

(d)
P(coin is at -2 after four moves) = 4C1 (7/12)^3 (5/12)

If the coin is at -2 after four moves, there is probability (7/12)^2 that the game will end on the sixth move.

P(coin is at 2 after four moves) = 4C1 (5/12)^3 (7/12)

If the coin is at 2 after four moves, there is probability (5/12)^2 that the game will end on the sixth move.

Thanks Johnny but I don't understand a part. Why do you use 4C1?

Jack

Question 15, Binomial Distribution:
B≈(24,1/36)
P(X>_1)
Therefore, using a binomial distribution table the probability is:
P(X>_1) = 0.4914038761

Question 12:
Define the events, P(A) = Lesserprod comp or not (Bestbits)
P(B) = Fail, or not fail
So 0.01 = (N*1/50))+((1-N)*1/200)
0.01 = 1/50N + 1/200 - 1/200N
0.01 - 1/200 = 4/200N - 1/200N
Therefore N = (0.01 - 1/200) / (3/200)=1/3
And this is the prop for Lesser so 1 - N=2/3 = Bestbits

Thanks Jack. But I don't understand why the p is 1/36? And for question 12, I don't really understand, why 0.01 = (N*1/50))+((1-N)*1/200)?

Thanks Zhang, Jonny W and Jack for helping me.

Reply 14

Jack

The reason p is 1/36 is because that is the probability of getting 2 6's on a dice.

(N*1/50))+((1-N)*1/200) - This is the formula for getting the probability to be 0.01. Define N as been the Probability of getting a Lesserprod (we don't know it so hence the usage of a letter), so the probability of getting a Bestbits is (1-N). And we already know the Probability of a component failing.

Reply 15

SlyPie

authecroix123

10)In a lottery there are 24 prizes allocated at random to 24 prize-
winners. Ann, Ben and Cal are three of the prize-winners. Of the
prizes, 4 are cars, 8 are bicycles and 12 are watches. Show that the
probability that Ann gets a car and Ben gets a bicycle or a watch is
10/69. Giving each answer either as a fraction or as a decimal correct
to 3 significant figures, find
a)the probability that both Ann and Ben get cars, given that Cal gets
a car,
b)the probability that either Ann or Cal(or both) gets a car,
c)the probability that Ann gets a car and Ben gets a car or a bicycle,
d)the probability that Ann gets a car given that Ben gets either a car
or a bicycle.

P(Ann gets a car and Ben gets a bicycle or watch)=10/69=
4/24 (8/23+12/23)= (4/24)(20/23)= 10/69
a) P (Ann and Ben get cars, given that Cal gets a car)=
(3/23)(2/22)

b) P(Ann gets car and Cal doesn't)+P(Cal gets car and Ann doesn't)+ P(both get a car)= (4/24)(20/23)+(4/24)(20/23)+(4/24)(3/23)= .312

c) P(Ann gets a car and Ben gets a car or a bicycle)= (4/24)(3/23+8/23)= .0797

Reply 16

SlyPie

d) P(Ann gets a car given that Ben gets either a car or a bicycle)
= P(Ann gets a car given Ben gets a car)+ P(ann gets a car given ben gets a bicycle)
= 3/23+ 4/23= 7/23

Reply 17

Jack

d) P(Ann gets a car given that Ben gets either a car or a bicycle)
= P(Ann gets a car given Ben gets a car)+ P(ann gets a car given ben gets a bicycle)
= 3/23+ 4/23= 7/23

No its not it is:
(8/12*4/23) + (4/12*3/23) = 11/69

Reply 18

SlyPie

Jack

No its not it is:
(8/12*4/23) + (4/12*3/23) = 11/69

Ok, I just read your correction wrong. It says GIVEN though, thus meaning it already happened. You do not have to take Ben into account.
--------------

authecroix123

5)The probability that an event A occurs is P(A)=0.3 .The event B is
independent of A and P(B)=0.4 .
a)Calculate P(A or B or both occur)
Event C is defined to be the event that neither A nor B occurs.
b)Calculate P(C|A'), where A' is the event that A does not occur.

Jack, are you ready?

a) P(A or B or both occur)
= (.3)(.6)+ (.4)(.7)+ (.4)(.3)=.58

b) P(C)= .6 since it's given A does not occur, then P(C)= P(that B does not occur)
--------------
Alright look, in the example of a stack of cards, the probably of drawing an ace GIVEN that one ace has already been drawn is simply 3/51 NOT (4/52)(3/51)

Jack, could you please further explain what you mean when you times out Ben given his chances? Por favor?

Reply 19

authecroix123

SlyPie

d) P(Ann gets a car given that Ben gets either a car or a bicycle)
= P(Ann gets a car given Ben gets a car)+ P(ann gets a car given ben gets a bicycle)
= 3/23+ 4/23= 7/23

Thanks SlyPie, I now understand how to do it now.