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S1 Probability Help (Edexcel)
1) An archer has two attempts to hit a target. The probability that his first arrow hits is 0.4 and the probability that the second arrow hits is 0.5. Given that he hits both targets with both arrows is 0.2, find the probability that he misses the target with both arrows.
2) If S and T are two events and P(T) =0.4, P(S n T) =0.15 and P(S' n T') =0.5, find:
(a) P(S n T') = 0.1
(b) P(S) = 0.25
(c) P(S u T) = 0.5
(d) P(S' n T) = 0.25
(e) P(S' u T') = ?
3) The events M and N are such that P(M) = P(N) = 2P(M n N). Given that P(M u N) = 0.6, find:
(a) P(M n N)
(b) P(M)
(c) P(M' n N')
(d) P(M n N')
If you could explain in words how you come to the answers aswell that'd be great!
Thank you
UPDATE MORE Q'S
5) Given P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6, find:
(a) P(R)
(b) P(S|R)
(c) P(S'|R)
(d) P(S'|R')
2) If S and T are two events and P(T) =0.4, P(S n T) =0.15 and P(S' n T') =0.5, find:
(a) P(S n T') = 0.1
(b) P(S) = 0.25
(c) P(S u T) = 0.5
(d) P(S' n T) = 0.25
(e) P(S' u T') = ?
3) The events M and N are such that P(M) = P(N) = 2P(M n N). Given that P(M u N) = 0.6, find:
(a) P(M n N)
(b) P(M)
(c) P(M' n N')
(d) P(M n N')
If you could explain in words how you come to the answers aswell that'd be great!
Thank you
UPDATE MORE Q'S
5) Given P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6, find:
(a) P(R)
(b) P(S|R)
(c) P(S'|R)
(d) P(S'|R')
Scroll to see replies
1) P(both miss)=P(first miss)*P(second miss)
Remember 1-hit=miss
2e) 1 if I'm reading it correctly... Because what is not in T is everything outside T and what is not in S is everything outside S.
3) P(M u N)=P(M)+P(N)-P(M n N)
This I think should get you started.
5) Use the formula booklet (I think it must be in there somewhere, Bayes theorem or some other formula dealing with conditional probability).
Remember 1-hit=miss
2e) 1 if I'm reading it correctly... Because what is not in T is everything outside T and what is not in S is everything outside S.
3) P(M u N)=P(M)+P(N)-P(M n N)
This I think should get you started.
5) Use the formula booklet (I think it must be in there somewhere, Bayes theorem or some other formula dealing with conditional probability).
Reply 2
nota bene
1) P(both miss)=P(first miss)*P(second miss)
Remember 1-hit=miss
2e) 1 if I'm reading it correctly... Because what is not in T is everything outside T and what is not in S is everything outside S.
3) P(M u N)=P(M)+P(N)-P(M n N)
This I think should get you started.
5) Use the formula booklet (I think it must be in there somewhere, Bayes theorem or some other formula dealing with conditional probability).
Remember 1-hit=miss
2e) 1 if I'm reading it correctly... Because what is not in T is everything outside T and what is not in S is everything outside S.
3) P(M u N)=P(M)+P(N)-P(M n N)
This I think should get you started.
5) Use the formula booklet (I think it must be in there somewhere, Bayes theorem or some other formula dealing with conditional probability).
Thanks that was really useful, is the formula the last one listed for probability?
Reply 4
hey i recognize those questions.. mind if i a question.. maybe more later...
Ex 5E book S1 Q1 if you have the book
1)
a bag A contains 4 yellow balls and 6 green balls and a similar bag B contains 5 yellow balls and 4 red balls. one ball is selected at random from A and then placed in B. one ball is then selected at random from B. let Yi, Gi, Ri, i = 1 , 2 represent the events that the ith ball selected is yellow, green or red respectively.
a) show that P(Y1 n R2) = 0.16
b) find the probability that the second ball selected is red
c) calculate P(G1|R2)
thanks!
edit :
another question
Ex 5E Q11 book S1 heinemann
the security passes for a certain company are coloured yellow or white and are provided with either a clip to go on a pocket or a chain to be worn around the neck. the probability that a pass has a clip is 6/10 and 2/3 of the white passes and 4/7 of the yellow ones are fitted with clips. A member of the company is stopped on his way into work and his pass is chkeced find the probability that
a) the pass is yellow
b) the pass is yellow with a chain.
if two people are stopped at random as they enter the company c) ifnd the probability that one pass will be yelow and the other white, and one will have a clip and the other a chain.
thanks!!
Ex 5E book S1 Q1 if you have the book
1)
a bag A contains 4 yellow balls and 6 green balls and a similar bag B contains 5 yellow balls and 4 red balls. one ball is selected at random from A and then placed in B. one ball is then selected at random from B. let Yi, Gi, Ri, i = 1 , 2 represent the events that the ith ball selected is yellow, green or red respectively.
a) show that P(Y1 n R2) = 0.16
b) find the probability that the second ball selected is red
c) calculate P(G1|R2)
thanks!
edit :
another question
Ex 5E Q11 book S1 heinemann
the security passes for a certain company are coloured yellow or white and are provided with either a clip to go on a pocket or a chain to be worn around the neck. the probability that a pass has a clip is 6/10 and 2/3 of the white passes and 4/7 of the yellow ones are fitted with clips. A member of the company is stopped on his way into work and his pass is chkeced find the probability that
a) the pass is yellow
b) the pass is yellow with a chain.
if two people are stopped at random as they enter the company c) ifnd the probability that one pass will be yelow and the other white, and one will have a clip and the other a chain.
thanks!!
Reply 5
Ex. 5E Q 1
P(Y1) = 0.4
P(R2) = 0.4
therefore P(Y1 n R2) =0 .4 x 0.4 = 0.16
b) Second ball red = 4/10
c) P/G1 n R2) = P (G1 n R2) / P(R2) = 0.6 x 0.4/0.4
Hope that helps.
P(Y1) = 0.4
P(R2) = 0.4
therefore P(Y1 n R2) =0 .4 x 0.4 = 0.16
b) Second ball red = 4/10
c) P/G1 n R2) = P (G1 n R2) / P(R2) = 0.6 x 0.4/0.4
Hope that helps.
Reply 6
shivmani
Ex. 5E Q 1
P(Y1) = 0.4
P(R2) = 0.4
therefore P(Y1 n R2) =0 .4 x 0.4 = 0.16
b) Second ball red = 4/10
c) P/G1 n R2) = P (G1 n R2) / P(R2) = 0.6 x 0.4/0.4
Hope that helps.
P(Y1) = 0.4
P(R2) = 0.4
therefore P(Y1 n R2) =0 .4 x 0.4 = 0.16
b) Second ball red = 4/10
c) P/G1 n R2) = P (G1 n R2) / P(R2) = 0.6 x 0.4/0.4
Hope that helps.
sorry i dont know if i am just stupid or very sleepy but how is p(y1) = 0.4 and p*r2) = 0.4?
thanks
Reply 7
probablity that you take a yellow one from the first bag is 4/10 = 0.4 and the second bag has 9 balls in all so when you place this one in the second bag, the total is 10.
Now the probability that you get a red one out is 4/10 = 0.4
The probability of the first event happening AND the second happening 0.4 X 0.4.
Now the probability that you get a red one out is 4/10 = 0.4
The probability of the first event happening AND the second happening 0.4 X 0.4.
Reply 8
hmmm
ok i thought the question meant us to find the probability that the 1st ball from the second bag was yellow and the second ball from the second bag was red..
no wonder they didnt say anything about replacement...
thanks ..
i feel stupid for misunderstand the question
ok i thought the question meant us to find the probability that the 1st ball from the second bag was yellow and the second ball from the second bag was red..
no wonder they didnt say anything about replacement...
thanks ..
i feel stupid for misunderstand the question
Reply 9
9) A and B are two independant events such that P(A) = α and P(A u B) = β, β > α. Show that:
P(B) = β - α / 1 - α
P(B) = β - α / 1 - α
Reply 10
For independent events
P(AUB) = P(A) + P(B) - P(A n B)
substitute the given probabilities in the above , bearing in mind that P(A n B) = P(A) x P(B) .
Rearrange, to get P(B) should be easy.
P(AUB) = P(A) + P(B) - P(A n B)
substitute the given probabilities in the above , bearing in mind that P(A n B) = P(A) x P(B) .
Rearrange, to get P(B) should be easy.
Reply 11
EX 5 E question 11
(Posted in this thread by hkchute)
Has anyone been able to do that yet? Can't seem to get the required answers
(Posted in this thread by hkchute)
Has anyone been able to do that yet? Can't seem to get the required answers
Reply 13
No, the answers are
a) 7/10
b) 3/10
c) 1/10
a) 7/10
b) 3/10
c) 1/10
Reply 14
yeh that question is confusing
Reply 15
there are probably easier ways to do it, but i got the right answers doing it like this (file attached)
hope it makes sense!
hope it makes sense!
Reply 16
this part of s1 is reallyyyyy mind boggling!!! can anyone explain to me how to do exercise 5D, Q5(a)(iii) of the heinemann s1 textbook please?? ty
ob.is.a.bit.good
9) A and B are two independant events such that P(A) = α and P(A u B) = β, β > α. Show that:
P(B) = β - α / 1 - α
P(B) = β - α / 1 - α
we know that:
P(A) = α
P(AuB) = β
we know they are independent, so:
P(AnB) = P(A)P(B)
hence the formula:
P(AuB) = P(A) + P(B) - P(AnB)
can be rearranged to:
P(AuB) = P(A) + P(B) - P(A)P(B)
putting the known values in:
β = α + P(B) + αP(B)
β - α = P(B)[1 - α]
P(B) = (β - α
/(1 - αAs for the chain and clip one...
Start with a tree diagram, ill try and describe it
1) The first "tree" has two branches, one Yellow and one White, call them y and w.
2) The second "tree" has four branches. From the w branch, there is a clip branch, 2/3, and a chain branch, 1/3. From the y branch, there is a clip branch, 4/7, and a chain branch, 3/7.
The next thing to do would be to fill in the unknown values on the tree diagram. You know that the sum of probabilities on each tree adds up to 1. So:
w + y = 1
You also are given that P(Clip) = 0.6. From this we can form the equation:
(2/3)w + (4/7)y = 6/10
Use these two simultaneous equations to achieve:
w = 3/10
y = 7/10
Now you can start the questions
1. P(Y) = 7/10, just worked this out
2. P(YnCh) = 7/10 x 3/7 = 3/10, from the tree diagram
3. There are two possibilites:
YCH WCL
YCL WCH
From the tree diagram you can easily figure out the probabilties of these to be:
P(YCHnWCL) = 3/50
P(YCLnWCH) = 1/25
3/50 + 1/25 = 1/10
Hope that helps
Start with a tree diagram, ill try and describe it
1) The first "tree" has two branches, one Yellow and one White, call them y and w.
2) The second "tree" has four branches. From the w branch, there is a clip branch, 2/3, and a chain branch, 1/3. From the y branch, there is a clip branch, 4/7, and a chain branch, 3/7.
The next thing to do would be to fill in the unknown values on the tree diagram. You know that the sum of probabilities on each tree adds up to 1. So:
w + y = 1
You also are given that P(Clip) = 0.6. From this we can form the equation:
(2/3)w + (4/7)y = 6/10
Use these two simultaneous equations to achieve:
w = 3/10
y = 7/10
Now you can start the questions
1. P(Y) = 7/10, just worked this out
2. P(YnCh) = 7/10 x 3/7 = 3/10, from the tree diagram
3. There are two possibilites:
YCH WCL
YCL WCH
From the tree diagram you can easily figure out the probabilties of these to be:
P(YCHnWCL) = 3/50
P(YCLnWCH) = 1/25
3/50 + 1/25 = 1/10
Hope that helps
Reply 19
Hi um i need help on a questions on probability...conditional probability T_T
7) C and D are two events and P(C|D)=1/3, P(C|D')=1/5 and P(D)=1/4, find..
a) P(CnD)
b) P(CnD')
c) P(C)
d) P(D|C)
e) P(D'|C)
f) P(D'|C')
Help needed...
7) C and D are two events and P(C|D)=1/3, P(C|D')=1/5 and P(D)=1/4, find..
a) P(CnD)
b) P(CnD')
c) P(C)
d) P(D|C)
e) P(D'|C)
f) P(D'|C')
Help needed...
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