M.Johnson2111
Badges: 9
Rep:
?
#1
Report Thread starter 3 weeks ago
#1
Hi,

Please can I have some help with this question.

I've found X = 0

But need help finding y and z please?
0
reply
M.Johnson2111
Badges: 9
Rep:
?
#2
Report Thread starter 3 weeks ago
#2
mmmmmmm
Attached files
0
reply
mqb2766
Badges: 18
Rep:
?
#3
Report 3 weeks ago
#3
What do you understand about independence for joint probabilities?
Last edited by mqb2766; 3 weeks ago
0
reply
M.Johnson2111
Badges: 9
Rep:
?
#4
Report Thread starter 3 weeks ago
#4
(Original post by mqb2766)
What do you understand about independence for joint probabilities?
If two events are independent of each other then the occurrence of one doesn't affect the probability of the occurrence of the other.

In this case I think this means the probability of event A doesn't affect the probability of the occurrence of event C.
0
reply
mqb2766
Badges: 18
Rep:
?
#5
Report 3 weeks ago
#5
(Original post by M.Johnson2111)
If two events are independent of each other then the occurrence of one doesn't affect the probability of the occurrence of the other.

In this case I think this means the probability of event A doesn't affect the probability of the occurrence of event C.
Sure, but what does it mean about how p(A and C) can be calculated from p(A) and p(C)?
0
reply
M.Johnson2111
Badges: 9
Rep:
?
#6
Report Thread starter 3 weeks ago
#6
(Original post by mqb2766)
Sure, but what does it mean about how p(A and C) can be calculated from p(A) and p(C)?
I'm not sure? Is the P (A) = P (C) ?
0
reply
mqb2766
Badges: 18
Rep:
?
#7
Report 3 weeks ago
#7
(Original post by M.Johnson2111)
I'm not sure? Is the P (A) = P (C) ?
No that simply means the two events have equal probability.
Do you not have the definition in your notes, it's very common?
https://www.mathsisfun.com/data/prob...dependent.html
Last edited by mqb2766; 3 weeks ago
1
reply
M.Johnson2111
Badges: 9
Rep:
?
#8
Report Thread starter 3 weeks ago
#8
(Original post by mqb2766)
No that simply means the two events have equal probability.
Do you not have the definition in your notes, it's very common?
https://www.mathsisfun.com/data/prob...dependent.html
No, I started the topic yesterday.

Thank you.

So P(A and C) = P(A) x P(B)
0
reply
M.Johnson2111
Badges: 9
Rep:
?
#9
Report Thread starter 3 weeks ago
#9
(Original post by mqb2766)
No that simply means the two events have equal probability.
Do you not have the definition in your notes, it's very common?
https://www.mathsisfun.com/data/prob...dependent.html
No, I started the topic yesterday.

Thank you.

So P(A and C) = P(A) x P(C)

Which would be:

(0.10 + y + z) x (0.39 + z)
0
reply
mqb2766
Badges: 18
Rep:
?
#10
Report 3 weeks ago
#10
(Original post by M.Johnson2111)
No, I started the topic yesterday.

Thank you.

So P(A and C) = P(A) x P(B)
For starting yesterday, it's a bit complex, but possible.
That's correct. Can you express that in terms of the two unknowns?
0
reply
mqb2766
Badges: 18
Rep:
?
#11
Report 3 weeks ago
#11
(Original post by M.Johnson2111)
(0.10 + y + z) x (0.39 + z)
What is the last expression equal to (in terms of the unknown?
0
reply
M.Johnson2111
Badges: 9
Rep:
?
#12
Report Thread starter 3 weeks ago
#12
(Original post by mqb2766)
No that simply means the two events have equal probability.
Do you not have the definition in your notes, it's very common?
https://www.mathsisfun.com/data/prob...dependent.html
No, I started the topic yesterday.

Thank you.

So P(A and C) = P(A) x P(C)

Which would be:

(0.10 + y + z) x (0.39 + z)

(Original post by mqb2766)
For starting yesterday, it's a bit complex, but possible.
That's correct. Can you express that in terms of the two unknowns?
By expanding the brackets I got:

Z^2 + 0.49z + zy + 0.039

This looks like a quadratic equation which I could solve but the zy confuses me. Would I take a factor of Z out of the expression?
0
reply
M.Johnson2111
Badges: 9
Rep:
?
#13
Report Thread starter 3 weeks ago
#13
By expanding the brackets I got:



Z^2 + 0.49z + zy + 0.039



This looks like a quadratic equation which I could solve but the zy confuses me. Would I take a factor of Z out of the expression?
0
reply
mqb2766
Badges: 18
Rep:
?
#14
Report 3 weeks ago
#14
It's not an equation. You've not put an = ?
Leave it factorized once you've made the equation.
0
reply
M.Johnson2111
Badges: 9
Rep:
?
#15
Report Thread starter 3 weeks ago
#15
(Original post by mqb2766)
It's not an equation. You've not put an = ?
Leave it factorized once you've made the equation.
Z (z + 0.49 + y) = 0.039
0
reply
mqb2766
Badges: 18
Rep:
?
#16
Report 3 weeks ago
#16
(Original post by M.Johnson2111)
Z (z + 0.49 + y) = 0.039
You can't just stick an = into an expression. You have
So P(A and C) = P(A) x P(C)
Which would be
? = (0.10 + y + z) (0.39 + z)

What is the variable representation of p(A and B)? That's what goes on the left hand side.
Last edited by mqb2766; 3 weeks ago
0
reply
M.Johnson2111
Badges: 9
Rep:
?
#17
Report Thread starter 3 weeks ago
#17
(Original post by mqb2766)
You can't just stick an = into an expression. You have
So P(A and C) = P(A) x P(C)
Which would be
? = (0.10 + y + z) x (0.39 + z)

What is the variable representation of p(A and B)? That's what goes on the left hand side.
By this do you mean:

P(A and C) = (0.10 + y + z) x (0.39 + z) ?
0
reply
mqb2766
Badges: 18
Rep:
?
#18
Report 3 weeks ago
#18
(Original post by M.Johnson2111)
By this do you mean:

P(A and C) = (0.10 + y + z) x (0.39 + z) ?
From the Venn diagram, which letter(s) or expression represents the p(A and C)?
You need an equation in terms of y and z to solve.
0
reply
M.Johnson2111
Badges: 9
Rep:
?
#19
Report Thread starter 3 weeks ago
#19
P (A n C) = z
0
reply
mqb2766
Badges: 18
Rep:
?
#20
Report 3 weeks ago
#20
(Original post by M.Johnson2111)
P (A n C) = z
Yes, so the equation is ... ?

You need another equation as you have two variables to determine. Can you think what that is?
https://www.ck12.org/probability/ven...bility-ALG-II/
It's not explicitly stated in the question, but it's a common property for probabilities.
Last edited by mqb2766; 3 weeks ago
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

What do you want most from university virtual open days and online events?

I want to be able to watch in my own time rather than turn up live (190)
29.41%
I want to hear more about the specifics of the course (106)
16.41%
I want to be able to dip in and dip out of lots of different sessions (58)
8.98%
I want to meet current students (54)
8.36%
I want to meet academics and the people that will be teaching me (51)
7.89%
I want to have a taster lecture or workshop to see what the teaching is like (128)
19.81%
My parents/guardians are more interested than me to be honest (38)
5.88%
Other things – I'll tell you in the thread (21)
3.25%

Watched Threads

View All
Latest
My Feed