# Statistics

#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.
0
2 years ago
#2
(Original post by 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.
Which part are you stuck with? Sometimes drawing venn diagrams helps for questions like this, so that could be a good place to start?
0
2 years ago
#3
(Original post by 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.
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.
Last edited by MarkFromWales; 2 years ago
0
#4
(Original post by 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?
Actually I couldn't understand what they were asking for. Thanks for the idea!
0
#5
(Original post by 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.
Thanks a lot. Didn't expect to get reply this quick. Thanks for that. What about the 2nd one? Any ideas?
0
2 years ago
#6
(Original post by Heidi002)
Thanks a lot. Didn't expect to get reply this quick. Thanks for that. What about the 2nd one? Any ideas?
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?
0
#7
(Original post by 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?
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.
0
2 years ago
#8
(Original post by 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.
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?
Last edited by mqb2766; 2 years ago
0
#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.
0
2 years ago
#10
(Original post by 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.
Again, some ideas pls, even if they're not fully correct?
0
#11
(Original post 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?
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?
0
#12
(Original post by mqb2766)
Again, some ideas pls, even if they're not fully?
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
Last edited by Heidi002; 2 years ago
0
2 years ago
#13
(Original post by 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?
There are 6 ways of choosing just 3 continents.
3 continents for the 1st choice
2 for the 2nd
1 for the 3rd.
So multiply 6/132 by 6.
0
2 years ago
#14
(Original post 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?
How do you get 5/132?
0
#15
(Original post by mqb2766)
How do you get 5/132?
I did mistake in that. It should be 1/22 if I put 3/12 × 4/11 × 5/10
0
2 years ago
#16
(Original post by Heidi002)
I did mistake in that. It should be 1/22 if I put 3/12 × 4/11 × 5/10
So multiply by 6. See #13.
0
2 years ago
#17
(Original post by 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 can't see how the answer is 20/83. Mistake in book?
0
2 years ago
#18
(Original post by 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
Agree with mqb2766

I get 4/19 too.
0
#19
(Original post by mqb2766)
So multiply by 6. See #13.
Ok. Thanks for spending much of your important time on the question.
0
#20
(Original post by mqb2766)
I can't see how the answer is 20/83. Mistake in book?
(Original post by ghostwalker)
Agree with mqb2766

I get 4/19 too.
Then my working is correct I guess.
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