# A-level maths probability- challenge question!!

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Hi if anyone can answer this question you're literally a genius:

I'm doing the new AS maths- it's edexcel and its from the stats and mechanics book

Here's the question:

What I don't get is that the answer is x= 5,7 and 9 but y must be even and must be bigger or equal to 20,

5 x 2 = 10 this alone is

Also, 5 x 5 = 25, which is bigger than 20 but is

thus, this doesn't satisfy the rule that p(y is even)=p(y>/=20)?

I'm doing the new AS maths- it's edexcel and its from the stats and mechanics book

Here's the question:

What I don't get is that the answer is x= 5,7 and 9 but y must be even and must be bigger or equal to 20,

5 x 2 = 10 this alone is

**smaller than 20**Also, 5 x 5 = 25, which is bigger than 20 but is

**not even**thus, this doesn't satisfy the rule that p(y is even)=p(y>/=20)?

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#3

(Original post by

Hi if anyone can answer this question you're literally a genius:

I'm doing the new AS maths- it's edexcel and its from the stats and mechanics book

Here's the question:

What I don't get is that the answer is x= 5,7 and 9 but y must be even and must be bigger or equal to 20,

5 x 2 = 10 this alone is

Also, 5 x 5 = 25, which is bigger than 20 but is

thus, this doesn't satisfy the rule that p(y is even)=p(y>/=20)?

**a_d2010**)Hi if anyone can answer this question you're literally a genius:

I'm doing the new AS maths- it's edexcel and its from the stats and mechanics book

Here's the question:

What I don't get is that the answer is x= 5,7 and 9 but y must be even and must be bigger or equal to 20,

5 x 2 = 10 this alone is

**smaller than 20**Also, 5 x 5 = 25, which is bigger than 20 but is

**not even**thus, this doesn't satisfy the rule that p(y is even)=p(y>/=20)?

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(Original post by

Y does not have to be bigger than 20 or even, but the probability that Y is even is the same as the probability is greater than or equal to 20.

**ecila21**)Y does not have to be bigger than 20 or even, but the probability that Y is even is the same as the probability is greater than or equal to 20.

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#7

(Original post by

That makes sense but I still don't get how they did it, why did they assume x must be odd?

**a_d2010**)That makes sense but I still don't get how they did it, why did they assume x must be odd?

2 x 4 = 8

7 x 4 = 28

5 x 4 = 20

2x

7x

5x

We know that the probability of Y being even is the same as the probability of Y being greater than or equal to 20. If x is even (Say, it's 2), then all of our values for Y will be even. Since 2 x 4 = 8, not all of our combinations will be greater than 20 so x must be odd. Sorry if that doesn't make perfect sense.

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#8

**a_d2010**)

Hi if anyone can answer this question you're literally a genius:

I'm doing the new AS maths- it's edexcel and its from the stats and mechanics book

Here's the question:

What I don't get is that the answer is x= 5,7 and 9 but y must be even and must be bigger or equal to 20,

5 x 2 = 10 this alone is

**smaller than 20**

Also, 5 x 5 = 25, which is bigger than 20 but is

**not even**

thus, this doesn't satisfy the rule that p(y is even)=p(y>/=20)?

Ok so I'll explain this as short as possible:

1) Find all possible values of Y ---> Y can be

**8,2x,28,7x,20 or 5x**

2) Find P(Y is even) --->

**P(Y is even) = P ( Y is 8,2x,28 or 20)**You know these are the values that must be even as in the question it says x is an integer but x cannot be even as otherwise the equality

**P(Y is even) = P(y>_ 20) wouldn't be true -**you can work this out because clearly if x is even

**P(Y is even) = 1**but

**P(Y>_20)**is clearly not 1 as

**8<20**so you can conclude x can't be even and must be odd -----> therefore 5x and 7x aren't even

3)

**P(combination of picks) = 1/2 * 1/3 = 1/6**Hence

**P(Y is even)**=

**4* 1/6 = 4/6 =2/3**you know this because there are six possible combinations of products you can have and 4 of them are even so

**4/6**combinations give even products.

**Therefore P(Y>_20) =2/3**

4) Now you know 2/3 of the product combinations must be greater than or equal to 20. Clearly 8 is smaller than 20 and 2x is the next smallest product- therefore

**28,7x,20 and 5x**must be products that are greater than 20

5) You get a system of inequalities ---->

**7x >_20, 5x>_20 , 2x<20**

**Solving these gives x>_20/7 , x>_4 and x<10 but x is an integer so x>_3, x>_4 and x<10**combing these inequalities gives the answer to the question that

**4<_x<10**

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(Original post by

If you write out all the possible combinations you get:

2 x 4 = 8

7 x 4 = 28

5 x 4 = 20

2x

7x

5x

We know that the probability of Y being even is the same as the probability of Y being greater than or equal to 20.

**ecila21**)If you write out all the possible combinations you get:

2 x 4 = 8

7 x 4 = 28

5 x 4 = 20

2x

7x

5x

We know that the probability of Y being even is the same as the probability of Y being greater than or equal to 20.

**If x is even (Say, it's 2), then all of our values for Y will be even. Since 2 x 4 = 8, not all of our combinations will be greater than 20 so x must be odd.**Sorry if that doesn't make perfect sense.**still**be 8, x being odd or even makes no difference to 2 x 4 as that will still be even.

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(Original post by

Ok so I'll explain this as short as possible:

1) Find all possible values of Y ---> Y can be

2) Find P(Y is even) --->

3)

4) Now you know 2/3 of the product combinations must be greater than or equal to 20. Clearly 8 is smaller than 20 and 2x is the next smallest product- therefore

5) You get a system of inequalities ---->

**Anonymouspsych**)Ok so I'll explain this as short as possible:

1) Find all possible values of Y ---> Y can be

**8,2x,28,7x,20 or 5x**2) Find P(Y is even) --->

**P(Y is even) = P ( Y is 8,2x,28 or 20)**You know these are the values that must be even as in the question it says x is an integer but x cannot be even as otherwise the equality**P(Y is even) = P(y>_ 20) wouldn't be true -**you can work this out because clearly**as**__if x is even P(Y is even) = 1 but P(Y>_20) is clearly not 1__**so you can conclude x can't be even and must be odd -----> therefore 5x and 7x aren't even**__8<20__3)

**P(combination of picks) = 1/2 * 1/3 = 1/6**Hence**P(Y is even)**=**4* 1/6 = 4/6 =2/3**you know this because there are six possible combinations of products you can have and 4 of them are even so**4/6**combinations give even products.**Therefore P(Y>_20) =2/3**4) Now you know 2/3 of the product combinations must be greater than or equal to 20. Clearly 8 is smaller than 20 and 2x is the next smallest product- therefore

**28,7x,20 and 5x**must be products that are greater than 205) You get a system of inequalities ---->

**7x >_20, 5x>_20 , 2x<20****Solving these gives x>_20/7 , x>_4 and x<10 but x is an integer so x>_3, x>_4 and x<10**combing these inequalities gives the answer to the question that**4<_x<10**
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#11

(Original post by

if x is odd, say 3 then 2 x 4 = 8 will

**a_d2010**)if x is odd, say 3 then 2 x 4 = 8 will

**still**be 8, x being odd or even makes no difference to 2 x 4 as that will still be even.**2 x 4 = 8**

**7 x 4 = 28**

**5 x 4 = 20**

2 x 2 = 4

7 x 2 = 14

5 x 2 = 10

As we can see, all of the solutions are even, meaning that P(Y is even) = 1. This should be equal to P(Y is greater than or equal to 20). However, one of the values for Y that stays the same regardless of the value of X is less than 20, meaning that P(Y is greater than or equal to 20) is not 1. This is the same for all even values of x, therefore x must be even. Is that slightly clearer?

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**Anonymouspsych**)

Ok so I'll explain this as short as possible:

1) Find all possible values of Y ---> Y can be

**8,2x,28,7x,20 or 5x**

2) Find P(Y is even) --->

**P(Y is even) = P ( Y is 8,2x,28 or 20)**You know these are the values that must be even as in the question it says x is an integer but x cannot be even as otherwise the equality

**P(Y is even) = P(y>_ 20) wouldn't be true -**you can work this out because clearly if x is even

**P(Y is even) = 1**but

__so you can conclude x can't be even and must be odd -----> therefore 5x and 7x aren't even__

**P(Y>_20)**is clearly not 1 as**8<20**3)

**P(combination of picks) = 1/2 * 1/3 = 1/6**Hence

**P(Y is even)**=

**4* 1/6 = 4/6 =2/3**you know this because there are six possible combinations of products you can have and 4 of them are even so

**4/6**combinations give even products.

**Therefore P(Y>_20) =2/3**

4) Now you know 2/3 of the product combinations must be greater than or equal to 20. Clearly 8 is smaller than 20 and 2x is the next smallest product- therefore

**28,7x,20 and 5x**must be products that are greater than 20

5) You get a system of inequalities ---->

**7x >_20, 5x>_20 , 2x<20**

**Solving these gives x>_20/7 , x>_4 and x<10 but x is an integer so x>_3, x>_4 and x<10**combing these inequalities gives the answer to the question that

**4<_x<10**

2/6=/=5/6 (they don't equal each other?)

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#13

(Original post by

sorry but as I said earlier what difference does x being odd make to 2 x 4 = 8, that will still be 8 nevertheless x being odd, that has no difference on it?

**a_d2010**)sorry but as I said earlier what difference does x being odd make to 2 x 4 = 8, that will still be 8 nevertheless x being odd, that has no difference on it?

**x**has absolutely no effect to

**Y being 8, 28 or 20**as x isn't even involved with these products

**.**It only affects the values of

**2x, 7x and 5x as x is embedded within the product.**So in other words

**x**only affects half of the product values.

Now the reason why

**x cannot be even**is that if it was, then

**2x, 7x and 5x would all be even**. And since

**8,28 and 20**are also all even this would mean every single possible product would be even if

**x was even**. However all the products can't be even otherwise

**P(Y is even)**would be

**1.**But the question says

**P(Y is even) = P(Y>_20) and P(Y>_20) is not 1 as 8 is obviously less than 20.**Therefore you

**must**deduce that

**x cannot be even.**

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(Original post by

Ok so first of all, the value of

Now the reason why

**Anonymouspsych**)Ok so first of all, the value of

**x**has absolutely no effect to**Y being 8, 28 or 20**as x isn't even involved with these products**.**It only affects the values of**2x, 7x and 5x as x is embedded within the product.**So in other words**x**only affects half of the product values.Now the reason why

**x cannot be even**is that if it was, then**2x, 7x and 5x would all be even**. And since**8,28 and 20**are also all even this would mean every single possible product would be even if**x was even**. However all the products can't be even otherwise**P(Y is even)**would be**1.**But the question says**P(Y is even) = P(Y>_20) and P(Y>_20) is not 1 as 8 is obviously less than 20.**Therefore you**must**deduce that**x cannot be even.**
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#15

(Original post by

Also, how would x being odd satisfy the p(y is even)=p(y>_20) because if x were odd then the probability of an odd combination would be 2/6 (7x and 5x) and the probability of y being bigger or equal to 20 would be 5/6 (28, 20, 2x, 7x and 5x)

2/6=/=5/6 (they don't equal each other?)

**a_d2010**)Also, how would x being odd satisfy the p(y is even)=p(y>_20) because if x were odd then the probability of an odd combination would be 2/6 (7x and 5x) and the probability of y being bigger or equal to 20 would be 5/6 (28, 20, 2x, 7x and 5x)

2/6=/=5/6 (they don't equal each other?)

**4<_x<10**, x can only take the values of

**5, 7 and 9**as x is also odd*

Now onto your question. You are getting confused with

**x being odd**and the whole product being

**odd.**If

**x is odd**then only even products would be

**8,20,28 and 2x**and

**cannot be 7x or 5x**. So 4/6 products would be even if x was odd. Secondly you are assuming that 2x is bigger than 20 but the idea of the question is to first figure out what

**P(Y is even)**is (which is 4/6) and then to use that information to work out the possible values of x. The question tells you that P(Y even) = P(Y>_20) but you have already worked out P(Y is even) = 4/6 (look above). This must mean that P(Y>_20) is also 4/6. And therefore 4/6 = 4/6

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#16

(Original post by

*Sorry from my previous inequality that

Now onto your question. You are getting confused with

**Anonymouspsych**)*Sorry from my previous inequality that

**4<_x<10**, x can only take the values of**5, 7 and 9**as x is also odd*Now onto your question. You are getting confused with

**x being odd**and the whole product being**odd.**If**x is odd**then only even products would be**8,20,28 and 2x**and**cannot be 7x or 5x**. So 4/6 products would be even if x was odd. Secondly you are assuming that 2x is bigger than 20 but the idea of the question is to first figure out what**P(Y is even)**is (which is 4/6) and then to use that information to work out the possible values of x. The question tells you that P(Y even) = P(Y>_20) but you have already worked out P(Y is even) = 4/6 (look above). This must mean that P(Y>_20) is also 4/6. And therefore 4/6 = 4/6
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#17

(Original post by

I still don't understand. Is there any other simpler way that you can explain this in?

**zh82916**)I still don't understand. Is there any other simpler way that you can explain this in?

1 ≥ P (Y is even) ≥ 4/6 since 8, 28, 20 & 2x are certainly even. But this cannot equal 5/6 because it is impossible for P (Y is even) to equal 5/6 as 7x and 5x would both have to be even if x is even (and hence the probability would be 1) and they would both be odd if x is odd (in which case the probability would be 4/6).

P (Y ≥ 20) ≤ 5/6 since 8 is the only value for Y that we certainly know is less than 20.

But clearly P(Y is even) = P (Y ≥ 20) from the question, so P(Y is even) = P (Y ≥ 20) = 4/6 from the inequalities and reasoning above.

This must mean x is odd so that 7x & 5x can't be even to retain P(Y is even) = 4/6. Hence, we need exactly 4 out of the 6 numbers for Y to be greater than 20 while x remains odd. x can't be 1 or 3 (otherwise less than 4 numbers for Y would be greater than 20) and x can't also be greater than 9 as otherwise more than 4 numbers for Y would be greater than 20.

**Hence x can be 5, 7 or 9 in this case.**

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