Hardest questions for Edexel C1, C3 and S1

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Oooooooooooooooooooooooooooooooooooooooo! Now I know my stupid mistake that shouldnt even be done!!!!! When I subtracted the gradent I got -10/5 which equals to 2!!!!!!! OMG, i hope every one learns from this just because you got a whole straight number doesnt mean that it's right. Thanx everyone. I feel so stupid. Common people, add any of the hardest questions you faced in C1,S1, and C3. Thanx alot. :smile:

Reply 21

robfinlay

No worries, once you get the methods and practice down its all a piece of cake. The achillies heel is tiny little mistakes in arithmetic and getting signs wrong, unfortunately something mostly everyone without exception does from time to time.

Hmm hard S1 question. Probability for me. (This is taken from Edexcel S1 Jan 2002).

Q4. A contractor bids for two building projects. He estimates that the probability of winning the first project is 0.5, and the probability of winning the second is 0.3 and the probability of winning both projects is 0.2.

a) Find the probability that he does not win either project.

b) Find the probability that he wins exactly one project.

c) Given that he does not win the first project, find the probability that he wins the second.

d) By calculation, determine whether or not winning the first contract and winning the second contract are independant events.

Probably an easy question but i cant do probability to save my life.

Cheers for any help,
Rob

Reply 22

ahmadz5

Ummm... Sorry man, it's 3:50 am here in amman :confused:

, and i've solved math to the point i got sick of it. This question is a bit tough. R u sure that every thing is correct? :confused:

Reply 23

robfinlay

Yeah all the info is correct. I dont have the mark scheme for that so i cant do a 'crafty' look up.

Reply 24

Dekota

robfinlay

Q4. A contractor bids for two building projects. He estimates that the probability of winning the first project is 0.5, and the probability of winning the second is 0.3 and the probability of winning both projects is 0.2.

P(A) = 0.5
P(B) = 0.3
P(A n B) = 0.2

a) Find the probability that he does not win either project.

P(A u B) = 0.2

b) Find the probability that he wins exactly one project.

=> (0.5 x 0.7) + (0.5 x 0.3)

Reply 25

ahmadz5

I wasnt thinking right! Ok, i thought that when he won both projects(0.2) does have in common which 0.5 and 0.3 but they dont cuz they mutally excluve becasue they didnt state that they were indepent!!! Draw it in a tree diagram.
So:........
"π" stands for intersection
(a) P(First project'π Second project')= 0.5x0.3=0.15
(b) P(one project)= (0.5x0.7)+(0.5x0.3)=0.5
(c) P(second|not first)= 0.15/0.5=0.3
(d) ???

Lsn i am not sure of these answers, if some one could verify. I am like 50% sure.

Reply 26

robfinlay

Cheers, dont know what you manipulated to get those answers though.

Thanks alot,
Rob

Reply 27

ahmadz5

V.P. Keys

P(A) = 0.5
P(B) = 0.3
P(A u B) = 0.2

P(A n B) = 0.2

=> (0.5 x 0.7) + (0.5 x 0.3)

=> (0.5 x 0.3)

V.P.Keys are you sure. Cuz P(A n B)= P(A)xP(B) = 0.15 not 0.2
Part (b) it is correct
but about part (c) why?

Reply 28

robfinlay

Hmm, understand a bit more. I cant check any answers as i dont have the mark scheme.

Reply 29

Dekota

ahmadz5

V.P.Keys are you sure. Cuz P(A n B)= P(A)xP(B) = 0.15 not 0.2
Part (b) it correct
but about part (d) why?

I wasn't too sure about them, you're probs right.

(d) is because you prove that P(A u B) is not the same as P(A u B) or something like that.

Reply 30

robfinlay

I got confused with the wording. I took either to mean none or both, p(does not win either) as being didnt win both of them.

Heh, this is why i dont like probability.

Reply 31

ahmadz5

What about part (c), shouldnt it be P(second|not first)= P(scond n not first)/p(not first)? and for (d) i think it's something bout intersection and union. Help please...

Reply 32

robfinlay

Ah!

if P(A|B) and P(A) are different, then events are not independant.
Also, if P(AnB) = P(A)P(B) events are independant.

Reply 33

Dekota

robfinlay

Ah!
http://www.thestudentroom.co.uk/search.php?do=getnew&1=2&3=4&5=6&7=8
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if P(A|B) and P(A) are different, then events are not independant.
Also, if P(AnB) = P(A)P(B) events are independant.

P(A) = 0.5
P(B) = 0.3
P(A n B) = 0.2

since 0.5 x 0.3 ≠ 0.2, the events are not independent?

Reply 34

ahmadz5

What about part (c)??? Could also some one check all the answers?

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V.P.Keys is ur calculater screwed cuz AxB= 0.15 not 0.2 and this is the second time u do it!