# Edexcel A Level Further Mathematics Paper 3B (9FM0 3B) - 14th June 2024 [Exam Chat]

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what are we assuming is an A* in this paper all?
Original post by Revi64
yeah i got this asw could someone confirm which is right?

it was greater than 5 not greater than or equal to im pretty sure. so P(X>=7) = 729/4096
Original post by MrMichi
what are we assuming is an A* in this paper all?

Probably around 55/75
Original post by seb9911
it was greater than 5 not greater than or equal to im pretty sure. so P(X>=7) = 729/4096

Yh but it would be 6-1 bcuz of what is said earlier no? Not 6+1
Original post by supergremlin
yh but it would be 6-1 bcuz of what is said earlier no? Not 6+1

p(x1+x2>5)
= p(y-1>5)
= p(y>6)
=p(y>=7)
Original post by Supergremlin
Q6d ) I got 81/256 for the probability of >=5
As Y = X1 + X2 +1
X1 +X2 = Y-1
Love to hear how you got 729/4096
it was bigger than, not bigger than or equal to (I think)
Original post by Supergremlin
Q6d ) I got 81/256 for the probability of >=5
As Y = X1 + X2 +1
X1 +X2 = Y-1
Love to hear how you got 729/4096

yh it I got 729/4096 and it was >5 not >=5. so I basically did y-1 = x1 + x2 and then the pgf of y-1 is 1/(4t+9) (I forgot the denominator) and then I expanded using binomial expansion to find the coefficients up to t^5. and then I did P(x1+x2 > 5) = 1-p(x<=5)
what did everyone get for the cell you had to calculate was it the one with frequency of 3 btw
what did everyone get for the cell you had to calculate was it the one with frequency of 3 btw

nah was bottom right as the expected frequency was the lowest
Original post by kettleboy
nah was bottom right as the expected frequency was the lowest

surely in context of the question he couldn’t have known that the bottom right has the lowest expected frequency?

Because he was selecting a cell in order to check the expected frequency only knowing the observed frequency, so surely 3 was right. that was my thinking anyway
Original post by Zacd25
surely in context of the question he couldn’t have known that the bottom right has the lowest expected frequency?
Because he was selecting a cell in order to check the expected frequency only knowing the observed frequency, so surely 3 was right. that was my thinking anyway

you choose the one with the lowest row total and the lowest column total which dont need maths to identify as expected = rowxcolumn / total
Original post by seb9911
p(x1+x2>5)
= p(y-1>5)
= p(y>6)
=p(y>=7)

Your right thank you for clarifying
literally i was on track for a good A* but from today I'm gonna scrape it by 5-10 marks
Original post by drake12334
literally i was on track for a good A* but from today I'm gonna scrape it by 5-10 marks
What mark you think you got overall?
Original post by sssdssss
What mark you think you got overall?

244 complete worse case
Original post by drake12334
244 complete worse case
Fingers crossed that’s enough for the A* then
Original post by Solysoly
I picked the one with 3 maybe there was a smaller one or another one with cell value 3?

I believe it was the bottom right cell, as that was the cell with the lowest probability. When working out Expected values.
Original post by kyan.cheung
Hi all, I've recorded the score distributions and my answers to each questions. Please tag me if you notice any issues!
1. (6 marks total)
a. (3 marks) Var(X)=4.9
b. (3 marks) Var(X^2)=97.69
2. (7 marks total)
ai. P(X=3) = 0.2090
aii. P(X<2)=0.3084 (2 marks total)
b. (4 marks) The Critical region is A<=1 or A>=12
c. (1 mark) P(Type I)=0.03744
3. (6 marks total)
a. (1 mark) Hypotheses - H0 colour and year group not associated, H1 colour and year group are associated
bi. Black Y10-12 because they are respectively the least frequent favourite colour and year group
ii. 5.383 (2 marks total)
c. (3 marks) (5-1)(3-1)=8 degrees of freedom with critical value 20.090, so reject H0 and conclude there is association between year group and favourite colour
4. (12 marks total)
a. (1 mark) X~Geo(1/6)
b. (2 marks) Show P(X<=3)=91/216
c. (5 marks) E[x]=6, Var(X)=30 -> Var(Xmean)30/64, P(5.6<Xmean<7.2)=0.6806
d. (4 marks) P(X>=16)=test value=0.064905...>0.05 so dont reject H0
5. (10 marks total)
a. (1 mark) H0 p=0.03 and H1 p>0.03
b. (2 marks) Size of test a 0.1821
c. (3 marks) Size of test b 0.2083
di. Power of a 0.7232
ii. We expect 94.877 compontents tested with test b (3 marks total)
e. Use test A as it has lower size and higher power (1 mark)
6. (16 marks total)
a. (6 marks) Variance 6
b. (4 marks) P(X<=2)=203/256
c. (4 marks) Y is a geometric distribution
d. (2 marks) P(X1+X2>5)=729/4096
7. (18 marks total)
a. (2 marks) Sam has prob 0.5886 of winning >=4 times
b. (4 marks) Don't remember what it asked but the answer was 0.04364 apparently
c. (6 marks) Show sam expects to win 1/25(n^2-16n). Sketch - basically set W~B(n, 1/5), then expected winnings is E[w^2+w-n] by sum rule. Then use E[w^2]=Var(W)+E^2[w]=4n/25+n^2/25 and simplify.
d. (4 marks) Sam has probability 0.3518 of not losing money
e. (4 marks) The other person has expected profit r^2-4r
RE predictions: with many people struggling on question 7, I think grade boundaries will be ~55 for A*.

bro why you’re even doing a levels ur in uni aren’t u😭😭