OCR Statistics 1- Probability question Watch

J DOT A
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Hey guys, I'm stuck and could really do with some help!

In a lottery there are 24 prizes allocated at random to 24 prize winners. Ann, Ben and Carl are three of the prize winners. Of these prizes, 4 are cars, 8 bikes and 12 watches.

i) find the probability that Ann gets a car and Ben gets a car or a bike.

So what I did was 4/24 * (3/23 + 8 /23) But thats not the correct answer, as the answer is 11/138? I dont see how they got that?

Help please
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ForGreatJustice
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4/24 * (3/23 + 8 /23)=11/138 unless i'm very much mistaken :P
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Maths_Lover
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(Original post by ForGreatJustice)
4/24 * (3/23 + 8 /23)=11/138 unless i'm very much mistaken :P
Damn, I was just about to make this comment but was beaten to it
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J DOT A
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(Original post by ForGreatJustice)
4/24 * (3/23 + 8 /23)=11/138 unless i'm very much mistaken :P
LMAO this has been a bad week for me. LOOL thank you again for coming to the rescue!

Okay here is an actual question I was stuck...

I)A simple model is that each child in any family is equally likely to be male or female, and that the sex of each child is independent of the sex of any previous children in familly. Using this model calculate the probability that, in a randomly chosen family of 4 children, there will be 2 males and 2 females.

So if there are 4 children, that means there are 8 possibilities. So surely its 1//2 to the power of 4 divided by 8? But the answer is 3/8?
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J DOT A
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(Original post by Maths_Lover)
Damn, I was just about to make this comment but was beaten to it
Okay here is an actual question I was stuck...

I)A simple model is that each child in any family is equally likely to be male or female, and that the sex of each child is independent of the sex of any previous children in familly. Using this model calculate the probability that, in a randomly chosen family of 4 children, there will be 2 males and 2 females.

So if there are 4 children, that means there are 8 possibilities. So surely its 1//2 to the power of 4 divided by 8? But the answer is 3/8?
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Maths_Lover
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(Original post by J DOT A)
Okay here is an actual question I was stuck...

I)A simple model is that each child in any family is equally likely to be male or female, and that the sex of each child is independent of the sex of any previous children in familly. Using this model calculate the probability that, in a randomly chosen family of 4 children, there will be 2 males and 2 females.

So if there are 4 children, that means there are 8 possibilities. So surely its 1//2 to the power of 4 divided by 8? But the answer is 3/8?
This is a binomial distribution question as there are 4 trials (the 4 children) and 2 different possibilities for each trial that are independent and unchanging ( 1/2 probability that a child is male and 1/2 probability that a child is female)

Therefore the answer is:

4C2 x (1/2)^2 x (1/2)^2
= 6 x 1/4 x 1/4
=3/8
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ForGreatJustice
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Isn't this just a binomial, though?

Given 4 children, what is the probability 2 of them are female?
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ForGreatJustice
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(Original post by Maths_Lover)
This is a binomial distribution question as there are 4 trials (the 4 children) and 2 different possibilities for each trial that are independent and unchanging ( 1/2 probability that a child is male and 1/2 probability that a child is female)

Therefore the answer is:

4C2 x (1/2)^2 x (1/2)^2
= 6 x 1/4 x 1/4
=3/8
beaten to it... if only my internet were faster...
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Maths_Lover
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(Original post by ForGreatJustice)
beaten to it... if only my internet were faster...
Pay back for earlier XD
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J DOT A
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(Original post by Maths_Lover)
Pay back for earlier XD
LOL Oh right I don't know how to do that yet so thats okay!

I have one more question, and ofc you or greatjustice don't need to do it as you have already helped me out alot!

i) Tests show that on average, 1 in 200 components produced by B fails within one year, and in in 200 components of L fail within one year of fitting. Given that 20% of components are B and 80% of components are L, what is the probability that if chosen at random, those chosen within a year of fitting will fail?

So What I did first was 20/100 * 1/200 =1/1000
then 80/100 * 1/50= 2/125
then multiply the 2 together... but the answer is 0.017
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J DOT A
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(Original post by ForGreatJustice)
beaten to it... if only my internet were faster...
LOL Oh right I don't know how to do that yet so thats okay!

I have one more question, and ofc you or greatjustice don't need to do it as you have already helped me out alot!

i) Tests show that on average, 1 in 200 components produced by B fails within one year, and one in 50 components of L fail within one year of fitting. Given that 20% of components are B and 80% of components are L, what is the probability that if chosen at random, those chosen within a year of fitting will fail?

So What I did first was 20/100 * 1/200 =1/1000
then 80/100 * 1/50= 2/125
then multiply the 2 together... but the answer is 0.017
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Maths_Lover
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(Original post by J DOT A)
LOL Oh right I don't know how to do that yet so thats okay!

I have one more question, and ofc you or greatjustice don't need to do it as you have already helped me out alot!

i) Tests show that on average, 1 in 200 components produced by B fails within one year, and in in 200 components of L fail within one year of fitting. Given that 20% of components are B and 80% of components are L, what is the probability that if chosen at random, those chosen within a year of fitting will fail?

So What I did first was 20/100 * 1/200 =1/1000
then 80/100 * 1/50= 2/125
then multiply the 2 together... but the answer is 0.017
??? in in 200?

Do not worry about it: I am happy to have been helpful
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J DOT A
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(Original post by Maths_Lover)
??? in in 200?

Do not worry about it: I am happy to have been helpful
i) Tests show that on average, 1 in 200 components produced by B fails within one year, and one in 50 components of L fail within one year of fitting. Given that 20% of components are B and 80% of components are L, what is the probability that if chosen at random, those chosen within a year of fitting will fail?



Sorry:P and thanks again!
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Maths_Lover
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(Original post by J DOT A)
i) Tests show that on average, 1 in 200 components produced by B fails within one year, and one in 50 components of L fail within one year of fitting. Given that 20% of components are B and 80% of components are L, what is the probability that if chosen at random, those chosen within a year of fitting will fail?



Sorry:P and thanks again!
That is OK, I was just momentarily confused

Right:

This is similar to calculating the weighted mean so what you would do is

(1/200 x 1/5) + (1/50 x 4/5)
=1/1000 + 2/125
=17/1000
=0.017
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J DOT A
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(Original post by Maths_Lover)
That is OK, I was just momentarily confused

Right:

This is similar to calculating the weighted mean so what you would do is

(1/200 x 1/5) + (1/50 x 4/5)
=1/1000 + 2/125
=17/1000
=0.017
wait why is it 1/200 x 1/5? and 1/50 x 4/5?
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Maths_Lover
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(Original post by J DOT A)
wait why is it 1/200 x 1/5? and 1/50 x 4/5?
1/200 = probability that B fails and you mutiply that by the percentage of total B components (1/5 = 20%) to get 1/1000.

1/50 = probability that L fails and you multiply this by the percentage of total L components (4/5 = 80%) to get 2/125.

At least this is what I think you mean: please tell me if it is not, I am quite bad at explaining things... :/

P.S. Sorry for the long wait, I was doing some dishes
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