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# S1 help. watch

1. the question is

probability of A and B are independent.

P(A)=2P(B) , P(AnB)= 0.125

show that P(B) = 0.25

My go,

i know that independent events allows you to show that,

P(A) x P(B) = P(AnB)

therefore, 0.125 = P(A) x 2P(B)
what next?
2. (Original post by Hi, How are you ?)
therefore, 0.125 = P(A) x P(B)
FTFY. From there, remember that you're given P(A) in terms of P(B).
3. (Original post by Hi, How are you ?)
the question is

probability of A and B are independent.

P(A)=2P(B) , P(AnB)= 0.125

show that P(B) = 0.25

My go,

i know that independent events allows you to show that,

P(A) x P(B) = P(AnB)

therefore, 0.125 = P(A) x 2P(B)
what next?
P(A) x P(B) = P(AnB)

First substitute P(AnB)=0.125:

P(A) x P(B) = 0.125

Now substitute P(A)=2P(B):

2P(B) x P(B) = 0.125

Can you continue from here?
4. (Original post by notnek)
P(A) x P(B) = P(AnB)

First substitute P(AnB)=0.125:

P(A) x P(B) = 0.125

Now substitute P(A)=2P(B):

2P(B) x P(B) = 0.125

Can you continue from here?
got the answers. but 1 quick q.

I got up to 2(P(B))2=0.125

how do i know whether to divide by 2 first and square root, or square root and divide by 2 then?
5. (Original post by Hi, How are you ?)
got the answers. but 1 quick q.

I got up to 2(P(B))2=0.125

how do i know whether to divide by 2 first and square root, or square root and divide by 2 then?
Always get rid of the constant first before you deal with the square root
6. (Original post by Hi, How are you ?)
got the answers. but 1 quick q.

I got up to 2(P(B))2=0.125

how do i know whether to divide by 2 first and square root, or square root and divide by 2 then?
Try to only square root if the whole of one of the sides is raised to a power of 2. So if you have

you should square root initially but if you have

you should remove the 2 first.
7. (Original post by cpdavis)
Always get rid of the constant first before you deal with the square root
I see, what rule dose that follow?
8. (Original post by Hi, How are you ?)
I see, what rule dose that follow?
SAMDIB

(BIDMAS in reverse)
9. (Original post by notnek)
SAMDIB

(BIDMAS in reverse)
For real?

thanks
10. (Original post by Hi, How are you ?)
I see, what rule dose that follow?
Just solving an equation in general.

So if I have:

I would divide by 2 so I have:

and solve for x.

Similarly, you have:

So what would you do? (Sorry if I come off as condescending )
11. Technically, you COULD square root first; it just wouldn't be as nice.

Thus, by the laws of surds:

Then proceed by dividing by sqrt(2) instead of 2.
12. (Original post by cpdavis)
Just solving an equation in general.

So if I have:

I would divide by 2 so I have:

and solve for x.

Similarly, you have:

So what would you do? (Sorry if I come off as condescending )
would you divide by 2, square root and the get the answers of rt5/4 ?
13. (Original post by Hi, How are you ?)
would you divide by 2, square root and the get the answers of rt5/4 ?
For my question yes (I realised that I forgot a 0 when typing it out )
14. (Original post by Hi, How are you ?)
would you divide by 2, square root and the get the answers of rt5/4 ?
Yup! But I think that was a typo on his part, and he wanted to link it back to the original question with .125 instead of a new question with .625 ^^
15. (Original post by cpdavis)
For my question yes (I realised that I forgot a 0 when typing it out )
sorry, what do you mean?
16. (Original post by aznkid66)
Yup! But I think that was a typo on his part, and he wanted to link it back to the original question with .125 instead of a new question with .625 ^^
sorry, what do you mean?
17. (Original post by Hi, How are you ?)
sorry, what do you mean?
I meant to type 0.0625, I accidentally typed 0.625 (If you do the calculation, you'll understand what I did )
18. Can someone help me with this question? I don't understand :c
The weight, X grams, of a particular variety of orange is normally distributed with mean 205
and standard deviation 25.
(a) Determine the probability that the weight of an orange is:
(i) less than 250 grams;

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