# Probability and Binomial Distribution Revision and Questions Help

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While revising probability I have found some questions to help improve my understanding. I have answered them fully and have shown my working I and was wondering if someone was able to advice whether I have used applicable methods to hopefully arrive at suitable conclusions or if my workings could be improved upon.

1. The probability of a hockey team winning a match is 0.6. Find the probability that:

a)They will not win one of the next three games

Probability of winning; P(W)=0.6

Probability of losing; P(W')=1-P(W)

P(W')=1-0.6=0.4

Since the events are mutually exclusive.

P(lose one game)=P(W⋂W⋂L) or P(W⋂L⋂W) or P(L⋂W⋂W)

P(lose one game)=(0.6*0.6*0.4)+(0.6*0.4*0.6)+(0.4*0.6*0.6)

P(lose one game)=0.144+0.144+0.144

P(lose one game)=0.432

b)They do not win their next three games

P(losing three games)=3P(W')=P(0.4⋂0.4⋂0.4)

3P(W')=0.064

c)They win the next five games in a row

P(win five games)=5P(W)=(0.6⋂0.6⋂0.6⋂0.6⋂0.6)

P(win five games)=0.07776

2. A bag contains 6 red balls, 4 white balls and 2 blue balls. Use this information to find the probability of choosing (without replacement);

a. Two balls of the same colour.

P(event)=number of favourable outcomes/total number of possible outcomes

P(red)=6/12=1/2

P(white)=4/12=1/3

P(blue)2/12=1/6

P(both same colour)=P(R⋂R) or P(W⋂W) or P(B⋂B)

P(both same colour)=(6/12*5/11)+(4/12+3/11)+(2/12+1/11)

P(both same colour)=5/22+1/11+1/66

P(both same colour)=1/3

b. Both of the blue balls;

P(blue)=2/12

P(both blue balls)=P(B⋂B)=P(2/12*1/11)

P(both blue balls)=1/66

If all of the balls are put back into the bag, then a further three balls are removed without replacement find the probability that;

c. The balls are all a different colour;

P(all a different colour)=P(R⋂W⋂B) or P(R⋂B⋂W) or P(W⋂R⋂B) or P(W⋂B⋂R) or P(B⋂R⋂W) or P(B⋂W⋂R)

P(all a different colour)=(6/12*4/11*2/10)+(6/12*2/11*4/10)+(4/12*6/11*2/10)+(4/12*2/11*6/10)+(2/12*6/11*4/10)+(2/12*4/11*6/10)

P(all a different colour)=2/55+2/55+2/55+2/55+2/55+2/55

P(all a different colour)=12/55

Or should I use Binomial distribution?

I am not sure how to put this in the form X~B(n,p) where n is the number of trials and p the probability of success.

I know that; P(X=x)=nCxp^x*q^n-x

So would that be here that n=12. I am not sure how to progress from here, or what p and q are?

I know that there are 12C3 ways to pick three balls from the bag.

If these three balls are all different colours, meaning one red, one white and one blue;

P(red)=6C1

P(white)=4C1

P(blue)=2C1

P(three different colours)=6C1*4C1*2C1/12C3

P(three different colours)=6*4*2/220

P(three different colours)=48/220

P(three different colours)=12/55

d. Find the probability there are more white balls than red balls;

There are more white balls if WWW are drawn or, WWR,WWB,WRW,WBB

P(more white than red)=(4/12*3/11*2/10)+(4/12*3/11*6/10)+(4/12*6/11*3/10)+(4/12*2/11*1/10)

P(more white than red)= 1/55+3/55+3/55+1/165

P(more white than red)=16/165

1. The probability of a hockey team winning a match is 0.6. Find the probability that:

a)They will not win one of the next three games

Probability of winning; P(W)=0.6

Probability of losing; P(W')=1-P(W)

P(W')=1-0.6=0.4

Since the events are mutually exclusive.

P(lose one game)=P(W⋂W⋂L) or P(W⋂L⋂W) or P(L⋂W⋂W)

P(lose one game)=(0.6*0.6*0.4)+(0.6*0.4*0.6)+(0.4*0.6*0.6)

P(lose one game)=0.144+0.144+0.144

P(lose one game)=0.432

b)They do not win their next three games

P(losing three games)=3P(W')=P(0.4⋂0.4⋂0.4)

3P(W')=0.064

c)They win the next five games in a row

P(win five games)=5P(W)=(0.6⋂0.6⋂0.6⋂0.6⋂0.6)

P(win five games)=0.07776

2. A bag contains 6 red balls, 4 white balls and 2 blue balls. Use this information to find the probability of choosing (without replacement);

a. Two balls of the same colour.

P(event)=number of favourable outcomes/total number of possible outcomes

P(red)=6/12=1/2

P(white)=4/12=1/3

P(blue)2/12=1/6

P(both same colour)=P(R⋂R) or P(W⋂W) or P(B⋂B)

P(both same colour)=(6/12*5/11)+(4/12+3/11)+(2/12+1/11)

P(both same colour)=5/22+1/11+1/66

P(both same colour)=1/3

b. Both of the blue balls;

P(blue)=2/12

P(both blue balls)=P(B⋂B)=P(2/12*1/11)

P(both blue balls)=1/66

If all of the balls are put back into the bag, then a further three balls are removed without replacement find the probability that;

c. The balls are all a different colour;

P(all a different colour)=P(R⋂W⋂B) or P(R⋂B⋂W) or P(W⋂R⋂B) or P(W⋂B⋂R) or P(B⋂R⋂W) or P(B⋂W⋂R)

P(all a different colour)=(6/12*4/11*2/10)+(6/12*2/11*4/10)+(4/12*6/11*2/10)+(4/12*2/11*6/10)+(2/12*6/11*4/10)+(2/12*4/11*6/10)

P(all a different colour)=2/55+2/55+2/55+2/55+2/55+2/55

P(all a different colour)=12/55

Or should I use Binomial distribution?

I am not sure how to put this in the form X~B(n,p) where n is the number of trials and p the probability of success.

I know that; P(X=x)=nCxp^x*q^n-x

So would that be here that n=12. I am not sure how to progress from here, or what p and q are?

I know that there are 12C3 ways to pick three balls from the bag.

If these three balls are all different colours, meaning one red, one white and one blue;

P(red)=6C1

P(white)=4C1

P(blue)=2C1

P(three different colours)=6C1*4C1*2C1/12C3

P(three different colours)=6*4*2/220

P(three different colours)=48/220

P(three different colours)=12/55

d. Find the probability there are more white balls than red balls;

There are more white balls if WWW are drawn or, WWR,WWB,WRW,WBB

P(more white than red)=(4/12*3/11*2/10)+(4/12*3/11*6/10)+(4/12*6/11*3/10)+(4/12*2/11*1/10)

P(more white than red)= 1/55+3/55+3/55+1/165

P(more white than red)=16/165

Last edited by LukeWatson4590; 1 year ago

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1. The probability of a hockey team winning a match is 0.6. Find the probability that:

a)They will not win one of the next three games

Probability of winning; P(W)=0.6

Probability of losing; P(W' )=1-P(W)

P(W' )=1-0.6=0.4

Since the events are mutually exclusive.

P(lose one game)=P(W⋂W⋂L) or P(W⋂L⋂W) or P(L⋂W⋂W)

P(lose one game)=(0.6*0.6*0.4)+(0.6*0.4*0.6)+(0.4*0.6*0.6)

P(lose one game)=0.144+0.144+0.144

P(lose one game)=0.432

**LukeWatson4590**)1. The probability of a hockey team winning a match is 0.6. Find the probability that:

a)They will not win one of the next three games

Probability of winning; P(W)=0.6

Probability of losing; P(W' )=1-P(W)

P(W' )=1-0.6=0.4

Since the events are mutually exclusive.

P(lose one game)=P(W⋂W⋂L) or P(W⋂L⋂W) or P(L⋂W⋂W)

P(lose one game)=(0.6*0.6*0.4)+(0.6*0.4*0.6)+(0.4*0.6*0.6)

P(lose one game)=0.144+0.144+0.144

P(lose one game)=0.432

For instance, for particular three matches, losing one match and winning two matches has a probability of 0.6^2 * 0.4.

In how many ways can you win and lose one game?

Well, you have 3 matches so just choose 1 of them to be lost. Hence there are 3C1 ways in which one match is lost and the other two are won.

Hence, the probability is (3C1) * 0.6^2 * 0.4.

This way of thinking would help you when you're considering bigger data; e.g. finding the probability of winning 40 out of 50 matches. Let's be honest, you're not going to sit there and draw as well as keep track of all the variations like you were able to on this smaller scale. Hence why the choose function becomes a necessity in wrapping sucha problem around your finger.

b)They do not win their next three games

P(losing three games)=3P(W' )=P(0.4⋂0.4⋂0.4)

3P(W' )=0.064

P(losing three games)=3P(W' )=P(0.4⋂0.4⋂0.4)

3P(W' )=0.064

You should write

Don't ever write -- this is a completely meaningless expression. The whole point is that P(A) means the probability of some event A occuring... but it P(0.4) is meaningless, what sense is it say that '0.4' is an event??

Also, don't write because this is not the same as which is what you mean!

c)They win the next five games in a row

P(win five games)=5P(W)=(0.6⋂0.6⋂0.6⋂0.6⋂0.6)

P(win five games)=0.07776

P(win five games)=5P(W)=(0.6⋂0.6⋂0.6⋂0.6⋂0.6)

P(win five games)=0.07776

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#3

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While revising probability I have found some questions to help improve my understanding. I have answered them fully and have shown my working I and was wondering if someone was able to advice whether I have used applicable methods to hopefully arrive at suitable conclusions or if my workings could be improved upon.

1. The probability of a hockey team winning a match is 0.6. Find the probability that:

a)They will not win one of the next three games

Probability of winning; P(W)=0.6

Probability of losing; P(W'=1-P(W)

P(W'=1-0.6=0.4

Since the events are mutually exclusive.

P(lose one game)=P(W⋂W⋂L) or P(W⋂L⋂W) or P(L⋂W⋂W)

P(lose one game)=(0.6*0.6*0.4)+(0.6*0.4*0.6)+(0.4*0.6*0.6)

P(lose one game)=0.144+0.144+0.144

P(lose one game)=0.432

**LukeWatson4590**)While revising probability I have found some questions to help improve my understanding. I have answered them fully and have shown my working I and was wondering if someone was able to advice whether I have used applicable methods to hopefully arrive at suitable conclusions or if my workings could be improved upon.

1. The probability of a hockey team winning a match is 0.6. Find the probability that:

a)They will not win one of the next three games

Probability of winning; P(W)=0.6

Probability of losing; P(W'=1-P(W)

P(W'=1-0.6=0.4

Since the events are mutually exclusive.

P(lose one game)=P(W⋂W⋂L) or P(W⋂L⋂W) or P(L⋂W⋂W)

P(lose one game)=(0.6*0.6*0.4)+(0.6*0.4*0.6)+(0.4*0.6*0.6)

P(lose one game)=0.144+0.144+0.144

P(lose one game)=0.432

b)They do not win their next three games

P(losing three games)=3P(W'=P(0.4⋂0.4⋂0.4)

3P(W'=0.064

P(losing three games) = P(W'nW'nW' = P(W'^3 = 0.4^3 = 0.064

P(win five games)=5P(W)=(0.6⋂0.6⋂0.6⋂0.6⋂0.6)

P(win five games)=0.07776

2. A bag contains 6 red balls, 4 white balls and 2 blue balls. Use this information to find the probability of choosing (without replacement);

a. Two balls of the same colour.

P(event)=number of favourable outcomes/total number of possible outcomes

P(red)=6/12=1/2

P(white)=4/12=1/3

P(blue)2/12=1/6

P(both same colour)=P(R⋂R) or P(W⋂W) or P(B⋂B)

P(both same colour)=(6/12*5/11)+(4/12+3/11)+(2/12+1/11)

P(both same colour)=5/22+1/11+1/66

P(both same colour)=1/3

a. Two balls of the same colour.

P(event)=number of favourable outcomes/total number of possible outcomes

P(red)=6/12=1/2

P(white)=4/12=1/3

P(blue)2/12=1/6

P(both same colour)=P(R⋂R) or P(W⋂W) or P(B⋂B)

P(both same colour)=(6/12*5/11)+(4/12+3/11)+(2/12+1/11)

P(both same colour)=5/22+1/11+1/66

P(both same colour)=1/3

P((R⋂R) u (W⋂W) u (B⋂B)) or you could say P((R⋂R) or (W⋂W) or (B⋂B))

=P(R⋂R) + P(W⋂W) + P(B⋂B)

b. Both of the blue balls;

P(blue)=2/12

P(both blue balls)=P(B⋂B)=P(2/12*1/11)

P(both blue balls)=1/66

If all of the balls are put back into the bag, then a further three balls are removed without replacement find the probability that;

c. The balls are all a different colour;

P(all a different colour)=P(R⋂W⋂B) or P(R⋂B⋂W) or P(W⋂R⋂B) or P(W⋂B⋂R) or P(B⋂R⋂W) or P(B⋂W⋂R)

P(all a different colour)=(6/12*4/11*2/10)+(6/12*2/11*4/10)+(4/12*6/11*2/10)+(4/12*2/11*6/10)+(2/12*6/11*4/10)+(2/12*4/11*6/10)

P(all a different colour)=2/55+2/55+2/55+2/55+2/55+2/55

P(all a different colour)=12/55

c. The balls are all a different colour;

P(all a different colour)=P(R⋂W⋂B) or P(R⋂B⋂W) or P(W⋂R⋂B) or P(W⋂B⋂R) or P(B⋂R⋂W) or P(B⋂W⋂R)

P(all a different colour)=(6/12*4/11*2/10)+(6/12*2/11*4/10)+(4/12*6/11*2/10)+(4/12*2/11*6/10)+(2/12*6/11*4/10)+(2/12*4/11*6/10)

P(all a different colour)=2/55+2/55+2/55+2/55+2/55+2/55

P(all a different colour)=12/55

Or should I use Binomial distribution?

I am not sure how to put this in the form X~B(n,p) where n is the number of trials and p the probability of success.

I know that; P(X=x)=nCxp^x*q^n-x

I am not sure how to put this in the form X~B(n,p) where n is the number of trials and p the probability of success.

I know that; P(X=x)=nCxp^x*q^n-x

So would that be here that n=12. I am not sure how to progress from here, or what p and q are?

I know that there are 12C3 ways to pick three balls from the bag.

If these three balls are all different colours, meaning one red, one white and one blue;

P(red)=6C1

P(white)=4C1

P(blue)=2C1

P(three different colours)=6C1*4C1*2C1/12C3

P(three different colours)=6*4*2/220

P(three different colours)=48/220

P(three different colours)=12/55

I know that there are 12C3 ways to pick three balls from the bag.

If these three balls are all different colours, meaning one red, one white and one blue;

P(red)=6C1

P(white)=4C1

P(blue)=2C1

P(three different colours)=6C1*4C1*2C1/12C3

P(three different colours)=6*4*2/220

P(three different colours)=48/220

P(three different colours)=12/55

d. Find the probability there are more white balls than red balls;

There are more white balls if WWW are drawn or, WWR,WWB,WRW,WBB

P(more white than red)=(4/12*3/11*2/10)+(4/12*3/11*6/10)+(4/12*6/11*3/10)+(4/12*2/11*1/10)

P(more white than red)= 1/55+3/55+3/55+1/165

P(more white than red)=16/165

There are more white balls if WWW are drawn or, WWR,WWB,WRW,WBB

P(more white than red)=(4/12*3/11*2/10)+(4/12*3/11*6/10)+(4/12*6/11*3/10)+(4/12*2/11*1/10)

P(more white than red)= 1/55+3/55+3/55+1/165

P(more white than red)=16/165

WWW, WWR, WRW, RWW, WWB, WBW, BWW, WBB.

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(Original post by

This is good. You can improve it by extending your idea of the fact that there are multiple ways in which one loss out of three games can occur, and thinking about it in terms of combinatorics.

For instance, for particular three matches, losing one match and winning two matches has a probability of 0.6^2 * 0.4.

In how many ways can you win and lose one game?

Well, you have 3 matches so just choose 1 of them to be lost. Hence there are 3C1 ways in which one match is lost and the other two are won.

Hence, the probability is (3C1) * 0.6^2 * 0.4.

This way of thinking would help you when you're considering bigger data; e.g. finding the probability of winning 40 out of 50 matches. Let's be honest, you're not going to sit there and draw as well as keep track of all the variations like you were able to on this smaller scale. Hence why the choose function becomes a necessity in wrapping sucha problem around your finger.

Notation is off here.

You should write

Don't ever write -- this is a completely meaningless expression. The whole point is that P(A) means the probability of some event A occuring... but it P(0.4) is meaningless, what sense is it say that '0.4' is an event??

Also, don't write because this is not the same as which is what you mean!

Same notational issues as above.

**RDKGames**)This is good. You can improve it by extending your idea of the fact that there are multiple ways in which one loss out of three games can occur, and thinking about it in terms of combinatorics.

For instance, for particular three matches, losing one match and winning two matches has a probability of 0.6^2 * 0.4.

In how many ways can you win and lose one game?

Well, you have 3 matches so just choose 1 of them to be lost. Hence there are 3C1 ways in which one match is lost and the other two are won.

Hence, the probability is (3C1) * 0.6^2 * 0.4.

This way of thinking would help you when you're considering bigger data; e.g. finding the probability of winning 40 out of 50 matches. Let's be honest, you're not going to sit there and draw as well as keep track of all the variations like you were able to on this smaller scale. Hence why the choose function becomes a necessity in wrapping sucha problem around your finger.

Notation is off here.

You should write

Don't ever write -- this is a completely meaningless expression. The whole point is that P(A) means the probability of some event A occuring... but it P(0.4) is meaningless, what sense is it say that '0.4' is an event??

Also, don't write because this is not the same as which is what you mean!

Same notational issues as above.

Oh yes, I see how I can improve my workings as you have shown.

e.g 1 a. the probability is (3C1) * 0.6^2 * 0.4

Could this be applied to finding when they lose three games and win five games?

Or should I keep my answers in their current format for b and c?

Yes, I fully suspected the notation was not correct I was just not sure how to express it correctly. Thank you very much for demonstrating that I greatly appreciate it. 👍

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(Original post by

Fine.

Know what you mean, but you want to say:

P(losing three games) = P(W'nW'nW' = P(W'^3 = 0.4^3 = 0.064

Same comment as previous.

Just correcting terminology.

P((R⋂R) u (W⋂W) u (B⋂B)) or you could say P((R⋂R) or (W⋂W) or (B⋂B))

=P(R⋂R) + P(W⋂W) + P(B⋂B)

Fine

Fine.

No, you don't want binomial as the situation doesn't meet the binomial's criteria.

Yes, that's another way to do it - nothing to do with binomial distribution as such.

There are more options here:

WWW, WWR, WRW, RWW, WWB, WBW, BWW, WBB.

**ghostwalker**)Fine.

Know what you mean, but you want to say:

P(losing three games) = P(W'nW'nW' = P(W'^3 = 0.4^3 = 0.064

Same comment as previous.

Just correcting terminology.

P((R⋂R) u (W⋂W) u (B⋂B)) or you could say P((R⋂R) or (W⋂W) or (B⋂B))

=P(R⋂R) + P(W⋂W) + P(B⋂B)

Fine

Fine.

No, you don't want binomial as the situation doesn't meet the binomial's criteria.

Yes, that's another way to do it - nothing to do with binomial distribution as such.

There are more options here:

WWW, WWR, WRW, RWW, WWB, WBW, BWW, WBB.

2d. Oh so if there are more options as you have shown;

Find the probability there are more white balls than red balls;

There are more white balls if WWW, WWR, WRW, RWW, WWB, WBW, BWW, WBB

P(more white than red)=(4/12*3/11*2/10)+(4/12*3/11*6/10)+(4/12*6/11*3/10)+(6/12*4/11*3/10)+(4/12*3/11*2/10)+(4/12*2/11*3/10)+(2/12*4/11*3/10)+(6/12*2/11*1/10)

P(more white than red)= 1/55+3/55+3/55+3/55+1/55+1/55+1/55+1/110

P(more white than red)=27/110

Are all of my others answers correct? Is there a faster method for the type of question like 2d instead of listing all of the possible orderings? 😁

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(Original post by

Thank you for your reply. I thought that the notation was not correct I was just not sure how to express it correctly. Thank you very much for demonstrating how to amend it.

2d. Oh so if there are more options as you have shown;

Find the probability there are more white balls than red balls;

There are more white balls if WWW, WWR, WRW, RWW, WWB, WBW, BWW, WBB

P(more white than red)=(4/12*3/11*2/10)+(4/12*3/11*6/10)+(4/12*6/11*3/10)+(6/12*4/11*3/10)+(4/12*3/11*2/10)+(4/12*2/11*3/10)+(2/12*4/11*3/10)+(6/12*2/11*1/10)

P(more white than red)= 1/55+3/55+3/55+3/55+1/55+1/55+1/55+1/110

P(more white than red)=27/110

Are all of my others answers correct? Is there a faster method for the type of question like 2d instead of listing all of the possible orderings? 😁

**LukeWatson4590**)Thank you for your reply. I thought that the notation was not correct I was just not sure how to express it correctly. Thank you very much for demonstrating how to amend it.

2d. Oh so if there are more options as you have shown;

Find the probability there are more white balls than red balls;

There are more white balls if WWW, WWR, WRW, RWW, WWB, WBW, BWW, WBB

P(more white than red)=(4/12*3/11*2/10)+(4/12*3/11*6/10)+(4/12*6/11*3/10)+(6/12*4/11*3/10)+(4/12*3/11*2/10)+(4/12*2/11*3/10)+(2/12*4/11*3/10)+(6/12*2/11*1/10)

P(more white than red)= 1/55+3/55+3/55+3/55+1/55+1/55+1/55+1/110

P(more white than red)=27/110

Are all of my others answers correct? Is there a faster method for the type of question like 2d instead of listing all of the possible orderings? 😁

You could use combinatorics as RDKGames suggested; this would reduce the amount of computation, and also allow you to combine some cases.

So, you'd have W,W,W as one case, W,W,W' as another case where R and B are combined to make W', and W,B,B as the final case.

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