Can anyone explain this probability question?

[ndearwater]

Q:
A pair of dice is rolled. A possible event is rolling a multiple of 5. What is the probability of the complement of this event?

The answer is: 29/36

That means that the probability of the event happening is 7/(6*6)
But I don't understand how the number of desired outcomes is 7.
A multiple of 5 would be: 5 and 10. But from there I'm lost on coming up with 7.

ALSO..... how about this one?

Q: Find the sum of the roots of the equation:

(Square root of [x-1]) + (Square root of [2x-1]) = x

Answer: x= 1, 5 so the sum is 6
I think the algebra is tripping me up or something but I cannot figure this out for the life of me.
Thanks guys!

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[nerdystar]

I'm not very good at explaining stuff, but I'll try...

The only multiples of 5 you can roll are 5 and 10. So the possible dice pairs you can roll are:

1 and 4
2 and 3
4 and 6
5 and 5

The chance of rolling each one is 1/6, but leave out the 1/6 for the reappearing # rolls - only count the 4 roll once (from the rolls of 1 and 4 and 4 and 6. This should give you 7/6 which you would then multiply by 1/6 for the probablilty of getting any one of these numbers in a roll. That should give you 7/36. I hope you could understand that! My method of explanation is not very clear. Anyone else feel free come chime in and clarify.

 

[KUMoose]

What's likely tripping you up is that rolling say
1 : 4
is different then
4 : 1 because each test has two independent trials (i.e. rolling die 1 has no impact on the outcome of die 2).

So the successes are
1 - 1 : 4
2 - 2 : 3
3 - 4 : 6
4 - 5 : 5
5 - 6 : 4
6 - 3 : 2
7 - 4 : 1

So that's 7 successes over 6^2 possibilities. The compliment is 1 - P(X), the probability of the event or 29/36.

 

[CuRy]

1) “A possible event is rolling a multiple of 5” means that sum of a pair dice is equal 5 or 10.
These are 7 possibilities: (1, 4) (2, 3) (3, 2) (4, 1) (4, 6) (5, 5) and (6, 4).

The complement is 6*6 -7 = 29. Its probability is 29/6*6 = 29/36. That’s it!

2) sqr(x-1) + sqr(2x-1) = x
Take square both sides then
(x-1) + 2*sqr(x-1)(2x-1) + (2x-1) = x^2
3x-2 + 2*sqr(x-1)(2x-1) = x^2
2*sqr(x-1)(2x-1) = x^2 -3x+2
2*sqr(x-1)(2x-1) = (x-1)(x-2)
Both sides have a MULTIPLE of sqr(x-1), so we have x=1 as one solution

2*sqr(2x-1) = [sqr(x-1)]*(x-2)
Take square both sides then
4(2x-1) = (x-1)*(x-2)^2
8x-4 = (x-1)*(x^2-4x+4)
8x-4 = x^3-4x^2+4x-x^2 +4x-4
8x-4 = x^3-5x^2+8x-4
0 = x^3-5x^2
0 = x^2*(x-5)
x=0 and x=5

Check : if x=0 then the original equation becomes sqr(-1) + sqr (-1) = 0: nonsense. Therefore, x=0 is eliminated.

The remaining solutions is x=1 and x=5.

Note: It is very time comsuming, so I don't think it will be in a real PCAT test! Don't worry!!!

 

[nerdystar]

Thanks for clarifying y'all! I was waayyy off! lol

 

[ndearwater]

Thank you guys so much! I am clear on the probability for sure

CURY,
Thanks for your explanation on the second problem. I'm following it pretty well but I'm just gonna copy/paste this section I don't understand:

A)2*sqr(x-1)(2x-1) = (x-1)(x-2)
Both sides have a MULTIPLE of sqr(x-1), so we have x=1 as one solution

B)2*sqr(2x-1) = [sqr(x-1)]*(x-2)

Okay, from the point where you say that both sides have a multiple of sqr(x-1)... what did you do? How did you get from the first equation (which I labeled A) to the second equation (B)? From that point on I follow it... Thank you for that detailed explanation!
I'm sure you're right about the PCAT not having such a time consuming problem. Math is definitley my worst area. I'm freaking out just a little

 

[CuRy]

CURY,
Thanks for your explanation on the second problem. I'm following it pretty well but I'm just gonna copy/paste this section I don't understand:

A)2*sqr(x-1)(2x-1) = (x-1)(x-2)
Both sides have a MULTIPLE of sqr(x-1), so we have x=1 as one solution

B)2*sqr(2x-1) = [sqr(x-1)]*(x-2)

Okay, from the point where you say that both sides have a multiple of sqr(x-1)... what did you do? How did you get from the first equation (which I labeled A) to the second equation (B)? From that point on I follow it... Thank you for that detailed explanation!
I'm sure you're right about the PCAT not having such a time consuming problem. Math is definitley my worst area. I'm freaking out just a little

A)2*sqr(x-1)(2x-1) = (x-1)(x-2)
Both sides have a MULTIPLE of sqr(x-1), so we have x=1 as one solution
[as I wrote sqr(x-1)(2x-1), meaning that sqr(x-1)*sqr(2x-1). The remainings are still correct!]

i.e. that
2*sqr(x-1)*sqr(2x-1) = (x-1)(x-2)
2*sqr(x-1)*sqr(2x-1) = [sqr(x-1)]^2 * (x-2)
Both sides have the same sqr(x-1), so
[sqrt(x-1)]*{2*sqr(2x-1) - [sqr(x-1)]*(x-2)} = 0
then each factor is equal 0
i.e.:
sqr(x-1) = 0 --> x=1 (as one solution)

and 2*sqr(2x-1) - [sqr(x-1)]*(x-2) = 0
--> B) 2*sqr(2x-1) = [sqr(x-1)]*(x-2)