# Statistics

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Heidi002

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#1

How to do this question?

The probability that the mother wins a prize is 1/6 and the probability that her daughter wins a prize is 2/7.

Assuming that the two events are independent find the probability that

a) either the mother or the daughter but not both wins the prize.

And another......

In a group of 12 international referees there are three from Africa, four from Asia and five from Europe. To officiate at a tournament three referees are chosen at random from the group. Calculate the probability that

a) A referee is chosen from each continent.

b) Exactly two referees are chosen from Asia.

Quick help is very much appreciated. Pre mocks starting this Monday.

The probability that the mother wins a prize is 1/6 and the probability that her daughter wins a prize is 2/7.

Assuming that the two events are independent find the probability that

a) either the mother or the daughter but not both wins the prize.

And another......

In a group of 12 international referees there are three from Africa, four from Asia and five from Europe. To officiate at a tournament three referees are chosen at random from the group. Calculate the probability that

a) A referee is chosen from each continent.

b) Exactly two referees are chosen from Asia.

Quick help is very much appreciated. Pre mocks starting this Monday.

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mqb2766

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#2

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#2

(Original post by

How to do this question?

The probability that the mother wins a prize is 1/6 and the probability that her daughter wins a prize is 2/7.

Assuming that the two events are independent find the probability that

a) either the mother or the daughter but not both wins the prize.

And another......

In a group of 12 international referees there are three from Africa, four from Asia and five from Europe. To officiate at a tournament three referees are chosen at random from the group. Calculate the probability that

a) A referee is chosen from each continent.

b) Exactly two referees are chosen from Asia.

Quick help is very much appreciated. Pre mocks starting this Monday.

**Heidi002**)How to do this question?

The probability that the mother wins a prize is 1/6 and the probability that her daughter wins a prize is 2/7.

Assuming that the two events are independent find the probability that

a) either the mother or the daughter but not both wins the prize.

And another......

In a group of 12 international referees there are three from Africa, four from Asia and five from Europe. To officiate at a tournament three referees are chosen at random from the group. Calculate the probability that

a) A referee is chosen from each continent.

b) Exactly two referees are chosen from Asia.

Quick help is very much appreciated. Pre mocks starting this Monday.

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MarkFromWales

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#3

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#3

(Original post by

How to do this question?

The probability that the mother wins a prize is 1/6 and the probability that her daughter wins a prize is 2/7.

Assuming that the two events are independent find the probability that

a) either the mother or the daughter but not both wins the prize.

**Heidi002**)How to do this question?

The probability that the mother wins a prize is 1/6 and the probability that her daughter wins a prize is 2/7.

Assuming that the two events are independent find the probability that

a) either the mother or the daughter but not both wins the prize.

First calculate P(M or D) = P(M) + P(D) - P(M and D)

Its says that M and D are independent events so P(M and D) = P(M) x P(D).

This answer still includes the probability that they both win the prize so next you need to subtract P(M and D).

As usual, drawing a Venn diagram can help you to understand questions like this.

Last edited by MarkFromWales; 2 years ago

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Heidi002

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#4

(Original post by

Which part are you stuck with? Sometimes drawing venn diagrams helps for questions like this, so that could be a good place to start?

**mqb2766**)Which part are you stuck with? Sometimes drawing venn diagrams helps for questions like this, so that could be a good place to start?

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Heidi002

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#5

(Original post by

Let M be the event, 'the mother wins a prize' and D the event, 'the daughter wins a prize'.

First calculate P(M or D) = P(M) + P(D) - P(M and D)

Its says that M and D are independent events so P(M and D) = P(M) x P(D).

This answer still includes the probability that they both win the prize so next you need to subtract P(M and D).

As usual, drawing a Venn diagram can help you to understand questions like this.

**MarkFromWales**)Let M be the event, 'the mother wins a prize' and D the event, 'the daughter wins a prize'.

First calculate P(M or D) = P(M) + P(D) - P(M and D)

Its says that M and D are independent events so P(M and D) = P(M) x P(D).

This answer still includes the probability that they both win the prize so next you need to subtract P(M and D).

As usual, drawing a Venn diagram can help you to understand questions like this.

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mqb2766

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#6

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#6

(Original post by

Thanks a lot. Didn't expect to get reply this quick. Thanks for that. What about the 2nd one? Any ideas?

**Heidi002**)Thanks a lot. Didn't expect to get reply this quick. Thanks for that. What about the 2nd one? Any ideas?

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#7

(Original post by

What are you stuck with? You should be able to set up the basic probabilities (come from each continent) and then think about how they combine? Pls post some ideas?

**mqb2766**)What are you stuck with? You should be able to set up the basic probabilities (come from each continent) and then think about how they combine? Pls post some ideas?

And for part b I am solving P(AsiaAsiaAfrica) + P(AsiaAsiaEurope) but the answer comes wrong.

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mqb2766

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#8

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#8

(Original post by

Look I'm using P(AfricaAsiaEurope) = 3/12 × 4/11 × 5/12 = 5/132 whereas the answer is 3/11 for part a

And for part b I am solving P(AsiaAsiaAfrica) + P(AsiaAsiaEurope) but the answer comes wrong.

**Heidi002**)Look I'm using P(AfricaAsiaEurope) = 3/12 × 4/11 × 5/12 = 5/132 whereas the answer is 3/11 for part a

And for part b I am solving P(AsiaAsiaAfrica) + P(AsiaAsiaEurope) but the answer comes wrong.

* Why do the denominators go 12, 11 and 12? How many remain after one is chosen? After 2 are chosen?

* When you select the first continent, how many continents could it be? How many for the 2nd, once the first is chosen? So what should you multply your basic probability you've just calculated by?

Last edited by mqb2766; 2 years ago

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#9

Another one

A box contains 20 chocolates of which 15 have soft centres and 5 have hard centres. Two chocolates are taken at random, one after the other. Calculate the probability that both chocolates have hard centres, given that the second chocolate has a hard centre.

A box contains 20 chocolates of which 15 have soft centres and 5 have hard centres. Two chocolates are taken at random, one after the other. Calculate the probability that both chocolates have hard centres, given that the second chocolate has a hard centre.

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#10

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#10

(Original post by

Another one

A box contains 20 chocolates of which 15 have soft centres and 5 have hard centres. Two chocolates are taken at random, one after the other. Calculate the probability that both chocolates have hard centres, given that the second chocolate has a hard centre.

**Heidi002**)Another one

A box contains 20 chocolates of which 15 have soft centres and 5 have hard centres. Two chocolates are taken at random, one after the other. Calculate the probability that both chocolates have hard centres, given that the second chocolate has a hard centre.

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#11

(Original post by

Not a bad start. Couple of things:

* Why do the denominators go 12, 11 and 12? How many remain after one is chosen? After 2 are chosen?

* When you select the first continent, how many continents could it be? How many for the 2nd, once the first is chosen? So what should you multply your basic probability you've just calculated by?

**mqb2766**)Not a bad start. Couple of things:

* Why do the denominators go 12, 11 and 12? How many remain after one is chosen? After 2 are chosen?

* When you select the first continent, how many continents could it be? How many for the 2nd, once the first is chosen? So what should you multply your basic probability you've just calculated by?

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#12

(Original post by

Again, some ideas pls, even if they're not fully?

**mqb2766**)Again, some ideas pls, even if they're not fully?

Last edited by Heidi002; 2 years ago

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#13

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#13

(Original post by

Sorry for the typing mistake. It is 5/10. But still the answer doesn't match. The answer comes 49/44. And I can't understand your second point. Can u simplify it a bit please?

**Heidi002**)Sorry for the typing mistake. It is 5/10. But still the answer doesn't match. The answer comes 49/44. And I can't understand your second point. Can u simplify it a bit please?

3 continents for the 1st choice

2 for the 2nd

1 for the 3rd.

So multiply 6/132 by 6.

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#14

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#14

**mqb2766**)

Not a bad start. Couple of things:

* Why do the denominators go 12, 11 and 12? How many remain after one is chosen? After 2 are chosen?

* When you select the first continent, how many continents could it be? How many for the 2nd, once the first is chosen? So what should you multply your basic probability you've just calculated by?

Pls reply to the thread, not pm.

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#15

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#16

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#16

(Original post by

I did mistake in that. It should be 1/22 if I put 3/12 × 4/11 × 5/10

**Heidi002**)I did mistake in that. It should be 1/22 if I put 3/12 × 4/11 × 5/10

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#17

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#17

(Original post by

Ohkay. I'm doing P(HH)/P(SH)+P(HH) = (5/20 × 4/19) / ((15/20 × 5/19) + (5/20 × 4/19)) = 4/19 but the answer is 20/83

**Heidi002**)Ohkay. I'm doing P(HH)/P(SH)+P(HH) = (5/20 × 4/19) / ((15/20 × 5/19) + (5/20 × 4/19)) = 4/19 but the answer is 20/83

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ghostwalker

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#18

**Heidi002**)

Ohkay. I'm doing P(HH)/P(SH)+P(HH) = (5/20 × 4/19) / ((15/20 × 5/19) + (5/20 × 4/19)) = 4/19 but the answer is 20/83

I get 4/19 too.

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#19

(Original post by

So multiply by 6. See #13.

**mqb2766**)So multiply by 6. See #13.

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#20

(Original post by

I can't see how the answer is 20/83. Mistake in book?

**mqb2766**)I can't see how the answer is 20/83. Mistake in book?

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