My thread of questions

[khalil]

At a certain school it was found that 35% of the 500 boys and 40% of the 400 girls enjoyed
bushwalking. One student from the school is chosen at random. Let G represent the event
that the student is a girl, and B represent the event that the student enjoys bushwalking.

Find Pr(G)....I dont know how to draw a tree diagram that will have an end branch of B and G

[kamil9876]

(assuming that only boys and girls go to this school, ie: a third sex doesn't exist)

[shinny]

At a certain school it was found that 35% of the 500 boys and 40% of the 400 girls enjoyed
bushwalking. One student from the school is chosen at random. Let G represent the event
that the student is a girl, and B represent the event that the student enjoys bushwalking.

Find Pr(G)....I dont know how to draw a tree diagram that will have an end branch of B and G

You can ignore B for this question. It's just asking the probability of finding a girl; nothing regarding whether they enjoy bushwalking or not. If there's 500 guys, 400 girls, using simple probability, the chance of choosing a girl is

[khalil]

At a certain school it was found that 35% of the 500 boys and 40% of the 400 girls enjoyed
bushwalking. One student from the school is chosen at random. Let G represent the event
that the student is a girl, and B represent the event that the student enjoys bushwalking.

Find Pr(G)....I dont know how to draw a tree diagram that will have an end branch of B and G

You can ignore B for this question. It's just asking the probability of finding a girl; nothing regarding whether they enjoy bushwalking or not. If there's 500 guys, 400 girls, using simple probability, the chance of choosing a girl is

lol. That was simple. Thanks

[khalil]

Pr(A) = Pr(A|B) Pr(B) + Pr(A|B') Pr(B')

Does Pr(B')=Pr(A)?

Also, Pr(A|B)=Pr(B|A)?

[shinny]

Pr(A) = Pr(A|B) Pr(B) + Pr(A|B') Pr(B')

Draw up a Venn diagram and you'll get...

Does Pr(B')=Pr(A)?

No, not in a general case. Venn diagram it and you'll see. Remember that there's things outside of both A and B.

Also, Pr(A|B)=Pr(B|A)?

No. , while

But seriously, just go learn how to start using Venn diagrams and you can derive all of these yourself. You're not learning anything by asking us to prove every single one of these for you, as shown by your continual need to keep asking. Spend some time and learn how to do it yourself. I don't mind helping, but there's no point if you're not getting anything out of it.

[khalil]

When I mean does Pr(A)= P(B'), I meant it in this situation.
Because isnt this how we get the law of probability? If not, someone please draw another tree diagram.

[shinny]

Oh, then yeh it would in a binary case with both events being mutually exclusive.

[jaja]

first you have to assign which is player is positive, ill take katia as positive
so its 1 and -x

1 = Pr(1) + Pr(3) + Pr(5)
-x = Pr(0) + Pr(2) + Pr(4) or you can just do 1-(Pr(1) + Pr(3) + Pr(5))

now in order it to be fare mean must equal zero

so Pr(1) + Pr(3) + Pr(5) - x(Pr(0) + Pr(2) + Pr(4)) = 0

solve for x

i know it has been a long time but i forgot to ask...why does one have a negative value'?

[Flaming_Arrow]

first you have to assign which is player is positive, ill take katia as positive
so its 1 and -x

1 = Pr(1) + Pr(3) + Pr(5)
-x = Pr(0) + Pr(2) + Pr(4) or you can just do 1-(Pr(1) + Pr(3) + Pr(5))

now in order it to be fare mean must equal zero

so Pr(1) + Pr(3) + Pr(5) - x(Pr(0) + Pr(2) + Pr(4)) = 0

solve for x

i know it has been a long time but i forgot to ask...why does one have a negative value'?

because its in terms of the players perspective, that is when one player loses money when another wins

[khalil]

first you have to assign which is player is positive, ill take katia as positive
so its 1 and -x

1 = Pr(1) + Pr(3) + Pr(5)
-x = Pr(0) + Pr(2) + Pr(4) or you can just do 1-(Pr(1) + Pr(3) + Pr(5))

now in order it to be fare mean must equal zero

so Pr(1) + Pr(3) + Pr(5) - x(Pr(0) + Pr(2) + Pr(4)) = 0

solve for x

i know it has been a long time but i forgot to ask...why does one have a negative value'?

because its in terms of the players perspective, that is when one player loses money when another wins

I dont get it either! Please explain further!

[Flaming_Arrow]

first you have to assign which is player is positive, ill take katia as positive
so its 1 and -x

1 = Pr(1) + Pr(3) + Pr(5)
-x = Pr(0) + Pr(2) + Pr(4) or you can just do 1-(Pr(1) + Pr(3) + Pr(5))

now in order it to be fare mean must equal zero

so Pr(1) + Pr(3) + Pr(5) - x(Pr(0) + Pr(2) + Pr(4)) = 0

solve for x

i know it has been a long time but i forgot to ask...why does one have a negative value'?

because its in terms of the players perspective, that is when one player loses money when another wins

I dont get it either! Please explain further!

ok just say i was playing a game with a friend where i win $1 if if its heads and he wins $1 if its tails, if he wins the game, i lose the the game i lose $1, which means -1

[khalil]

Much better thanks. can someone help me out with this question. I teacher is terrible and cant answer it. PLEASE HELP.

A newspaper seller buys papers for 50 cents and sells them for 75 cents, and cannot return
unsold papers.

If the newspaper seller stocks too many papers a loss is incurred. If too few papers are
stocked potential profit is lost because of the excess demand. Let s represent the number
of newspapers stocked, and X the daily demand.

a) If P is the newspaper seller’s profit for a particular stock level s, find and expression
for P in terms of s and X.

ans. 0.75x-0.5s  x<s
       0.5s-0.25x  x>s

[TrueTears]

I remember asking about this question... if my memory serves me correct I think the answers were wrong...

Ok I've confirmed the book having the wrong answers.

[khalil]

but what are the right answers?...should i just avoid the question?