# A-level maths probability- challenge question!!

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Thread starter 2 years ago
#1
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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Thread starter 2 years ago
#2
why can't x be even because it satifies the rule
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2 years ago
#3
(Original post by 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)?
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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Thread starter 2 years ago
#4
(Original post by 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.
That makes sense but I still don't get how they did it, why did they assume x must be odd?
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Thread starter 2 years ago
#5
anyone...
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Thread starter 2 years ago
#6
...
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2 years ago
#7
(Original post by a_d2010)
That makes sense but I still don't get how they did it, why did they assume x must be odd?
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.
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2 years ago
#8
(Original post by 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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Thread starter 2 years ago
#9
(Original post by 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.
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.
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Thread starter 2 years ago
#10
(Original post by 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 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
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?
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2 years ago
#11
(Original post by 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.
Sorry, I wasn't very clear with that bit. If we say that x = 2, our combinations will be: (The ones in bold are the ones that stay the same regardless of the value of x)

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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Thread starter 2 years ago
#12
(Original post by 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 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
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?)
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2 years ago
#13
(Original post by 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?
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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Thread starter 2 years ago
#14
(Original post by 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.
I get what your saying now thank you so much x
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2 years ago
#15
(Original post by 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?)
*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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4 weeks ago
#16
(Original post by 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
I still don't understand. Is there any other simpler way that you can explain this in?
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2 weeks ago
#17
(Original post by zh82916)
I still don't understand. Is there any other simpler way that you can explain this in?
Y = { 8, 28, 20, 2x, 7x, 5x }

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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