Find how many different numbers can be made from some or all of the digits of the number 1345789

It had to be 57. I checked by physically writing down all possible numbers on paper lol.

The mid-points and mean have been updated. The ones mentioned previously were incorrect. Q1 [4 marks] n=19 probability of success=0.12, find fewer than 4 so apply binomial with 0,1,2 and 3

Ans: 0.813

Ans: 231

Ans: (i) 0.72 (ii) 0.417

Ans : (ii) 0: 3/36 1:13/36 2: 16/36 3:4/36 (iii) 1.58

Ans: (i) 720 (ii) 60 (iii) 57

Ans: (i) 6 (iii) 11.7, 0.547

Ans: (i) 0.653 (ii) 0.561 (iii) 0.321

Feel free to post any queries or mistakes

It is 60. I accidentally multiplied by 5 (please don't ask why, so many 5P something got the 5 stuck in my head and I subsconciously wrote down 5 instead of 3 and got 100), but yes, the answer is 60. It was actually one of the easier questions IMO. It had to be 60 because the paper clearly stated NO REPETITIONS for this question. 5P2 for each x 3 Good luck to everyone.

average though i messed up the easiest question

Q1 [4 marks] n=19 probability of success=0.12, find fewer than 4 so apply binomial with 0,1,2 and 3

Ans: 0.813

Q2 [4 marks] Group A was of 7, Group B was of 2, Group C was of 2. Choose 5 and at least 1 from each group. Possibilities: 3 from A and 1 from B and 1 from C. 7C3*2C1*2C1=140 2 from A and 2 from B and 1 from C. 7C2*2C2*2C1=42 2 from A and 1 from B and 2 from C. 7C2*2C1*2C2-42 1 from A and 2 from B and 2 from C. 7C1*2C2*2C2=7 Add them.

Ans: 231

Q3 (i) [3 marks] (ii) [2 marks] Roger had a probability of winning first set=0.6 Roger had a probability of winning second set if he had won the first set=0.7 Roger had a probability of winning second set if he lost the first set=0.25 Match would end when any player wins 2 sets and match cannot be drawn and Andy was the other player. (i) Probability that match would end in 2 sets. If Roger wins: 0.6*0.7=0.42 If Andy wins: 0.4*0.75=0.3 Add them. (ii) If match ends in 2 sets find the probability that match Andy won the match 0.3/0.72

Ans: (i) 0.72 (ii) 0.417

Q4 (i) [3 marks] (ii) [4 marks] (iii) [2 marks] Coin A is thrown twice and Coin B once Coin A has probability of head= 2/3 Coin B has probability of head= 1/4 (i) Show that the probability of getting 1 head is 13/36 If Coin A gives 1 head and Coin B gives none: 2*2/3*1/3*3/4=12/36 (Multiplied by 2 as both throws on Coin A can give a head) If Coin A gives 0 heads and Coin B gives 1 head: 1/3*1/3*1/4=1/36 Add them and total= 13/36 (ii) Draw a probability distribution table of numbers of heads obtained 0 when no heads obtained: 1/3*1/3*3/4=3/36 2 when Coin A gives 2 heads: 2/3*2/3*3/4=12/36 Both Coin A and Coin B give one head each: 2*1/3*2/3*1/4=4/36 (Multiplied by 2 as both throws on Coin A can give a head) Add them. 3 when all heads: 2/3*2/3:1/4=4/36 (iii) Find the expected value 1*13/36 + 2*16/36 +3*4/36=19/12

Ans : (ii) 0: 3/36 1:13/36 2: 16/36 3:4/36 (iii) 1.58

Q5 (i) [2 marks] (ii) [3 marks] (iii) [3 marks] 1345789 is a number (i) Find the number of arrangements when odd numbers together and digits not repeated 5!*3! =720 (ii) Arrangements between 3000 and 5000 and even numbers and digits not repeated When starting with 3: Last digit can be 4 or 8 so the middle two digits are 5P2=5P2*2!=40 When starting with 4: Last digit is 4 so middle two digits are 5P2=20 Total arrangements = 60 (iii) Multiples of 5 less than 1000 and digits ARE REPEATED Last digit has to be 5 and can be 3-digit, 2-digit or 1-digit number When 3 digit number: 7*7*1=49 When 2 digit number: 7*1=7 When 1-digit number: 1 Total =57

Ans: (i) 720 (ii) 60 (iii) 57

Q6 (i) [1 mark] (ii) [4 marks] (iii) [4 marks] Total athletes=57 <10 were 0 <10.5 were 4 <11.0 were 10 <12.0 were 40 <12.5 were 49 <13.5 were 57 (i) Find in the interval 10.5-11.0 =6 (ii) Draw a histogram Frequencies: 4,6,30,9,8 Frequency densities: 8,12,30,18,8 (iii) Calculate mean and variance Mid-points: 10.2, 10.7, 11.45, 12.2, 12.95 Mean=11.6 Variance=0.547

Ans: (i) 6 (iii) 11.6, 0.547

Q7 (i) [3 marks] (ii) [2 marks] (iii) [6 marks] Rafa studies at mean rate 1.9. The probability of studying more than 1.35 is 0.8 (i) Find Standard Deviation Z=0.842 so S.D. =0.653 (ii) Find the probability that studies less than 2.0 (2.0-1.9)/0.653=0.1531 P=0.561 (iii) 200 samples are taken. Find the probability between 163 and 173 inclusive np=160 npq=32 Less than 173.5: 0.9915 Less than 162.5: 0.6707 Between them= 0.321

Ans: (i) 0.653 (ii) 0.561 (iii) 0.321

Feel free to post any queries or mistakes

i gave variant 2 Kamihus what was your variant?

Yes!!! Get in!!!
I am so happy. According to this mark scheme, I got 46/50 (given I made no stupid errors I don't remember). Plus 69/75 on the other paper. I hope this secures me an A. So happy!!

Ho

Q1 [4 marks] n=19 probability of success=0.12, find fewer than 4 so apply binomial with 0,1,2 and 3

Ans: 0.813

Q2 [4 marks] Group A was of 7, Group B was of 2, Group C was of 2. Choose 5 and at least 1 from each group. Possibilities: 3 from A and 1 from B and 1 from C. 7C3*2C1*2C1=140 2 from A and 2 from B and 1 from C. 7C2*2C2*2C1=42 2 from A and 1 from B and 2 from C. 7C2*2C1*2C2-42 1 from A and 2 from B and 2 from C. 7C1*2C2*2C2=7 Add them.

Ans: 231

Q3 (i) [3 marks] (ii) [2 marks] Roger had a probability of winning first set=0.6 Roger had a probability of winning second set if he had won the first set=0.7 Roger had a probability of winning second set if he lost the first set=0.25 Match would end when any player wins 2 sets and match cannot be drawn and Andy was the other player. (i) Probability that match would end in 2 sets. If Roger wins: 0.6*0.7=0.42 If Andy wins: 0.4*0.75=0.3 Add them. (ii) If match ends in 2 sets find the probability that match Andy won the match 0.3/0.72

Ans: (i) 0.72 (ii) 0.417

Q4 (i) [3 marks] (ii) [4 marks] (iii) [2 marks] Coin A is thrown twice and Coin B once Coin A has probability of head= 2/3 Coin B has probability of head= 1/4 (i) Show that the probability of getting 1 head is 13/36 If Coin A gives 1 head and Coin B gives none: 2*2/3*1/3*3/4=12/36 (Multiplied by 2 as both throws on Coin A can give a head) If Coin A gives 0 heads and Coin B gives 1 head: 1/3*1/3*1/4=1/36 Add them and total= 13/36 (ii) Draw a probability distribution table of numbers of heads obtained 0 when no heads obtained: 1/3*1/3*3/4=3/36 2 when Coin A gives 2 heads: 2/3*2/3*3/4=12/36 Both Coin A and Coin B give one head each: 2*1/3*2/3*1/4=4/36 (Multiplied by 2 as both throws on Coin A can give a head) Add them. 3 when all heads: 2/3*2/3:1/4=4/36 (iii) Find the expected value 1*13/36 + 2*16/36 +3*4/36=19/12

Ans : (ii) 0: 3/36 1:13/36 2: 16/36 3:4/36 (iii) 1.58

Q5 (i) [2 marks] (ii) [3 marks] (iii) [3 marks] 1345789 is a number (i) Find the number of arrangements when odd numbers together and digits not repeated 5!*3! =720 (ii) Arrangements between 3000 and 5000 and even numbers and digits not repeated When starting with 3: Last digit can be 4 or 8 so the middle two digits are 5P2=5P2*2!=40 When starting with 4: Last digit is 4 so middle two digits are 5P2=20 Total arrangements = 60 (iii) Multiples of 5 less than 1000 and digits ARE REPEATED Last digit has to be 5 and can be 3-digit, 2-digit or 1-digit number When 3 digit number: 7*7*1=49 When 2 digit number: 7*1=7 When 1-digit number: 1 Total =57

Ans: (i) 720 (ii) 60 (iii) 57

Q6 (i) [1 mark] (ii) [4 marks] (iii) [4 marks] Total athletes=57 <10 were 0 <10.5 were 4 <11.0 were 10 <12.0 were 40 <12.5 were 49 <13.5 were 57 (i) Find in the interval 10.5-11.0 =6 (ii) Draw a histogram Frequencies: 4,6,30,9,8 Frequency densities: 8,12,30,18,8 (iii) Calculate mean and variance Mid-points: 10.2, 10.7, 11.45, 12.2, 12.95 Mean=11.6 Variance=0.547

Ans: (i) 6 (iii) 11.6, 0.547

Q7 (i) [3 marks] (ii) [2 marks] (iii) [6 marks] Rafa studies at mean rate 1.9. The probability of studying more than 1.35 is 0.8 (i) Find Standard Deviation Z=0.842 so S.D. =0.653 (ii) Find the probability that studies less than 2.0 (2.0-1.9)/0.653=0.1531 P=0.561 (iii) 200 samples are taken. Find the probability between 163 and 173 inclusive np=160 npq=32 Less than 173.5: 0.9915 Less than 162.5: 0.6707 Between them= 0.321

Ans: (i) 0.653 (ii) 0.561 (iii) 0.321

Feel free to post any queries or mistakes [/

Ho

Which formula did you use for the mean and variance in q6iii

The one for the grouped data. These are by using the exact values. I also solved them on the stat mode on the calculator and got this answer.

and your variance should be 0.600 your midpoints are not right.10.25,10.75,11.5 so on.

BTW, due to accuracy issues, it should have been 11.7 for the mean, not 11.6 (11.66 rounds up). You used 10.2 and 10.7, should have been 10.25 and 12.75. CIE clearly states you should work to 4 s.f.'s and give answers to 3 s.f.'s

And variance was 0.6! Once again ,due to accuracy!

Exactly. Thank you for stating that

BTW, due to accuracy issues, it should have been 11.7 for the mean, not 11.6 (11.66 rounds up). You used 10.2 and 10.7, should have been 10.25 and 12.75. CIE clearly states you should work to 4 s.f.'s and give answers to 3 s.f.'s

The

Yes mean will be calculated that way.

Check the Q3 (iii) for this paper. Less than was used here too.

I found this one easier. Although its better to wait for the actual thresholds but it surely will be 35-40 range

.

BTW, due to accuracy issues, it should have been 11.7 for the mean, not 11.6 (11.66 rounds up). You used 10.2 and 10.7, should have been 10.25 and 12.75. CIE clearly states you should work to 4 s.f.'s and give answers to 3 s.f.'s

The method you mentioned was correct for the mean. My mean and variance are correct for MY mid-points.

BTW For the histograms, did you ensure the line did not touch exactly the line for the boundaries for the right? As in <12 so being at like 11.95

I got the same midpoints and frequencies as you but I seem to get a different answer , that's what's confusing me I thought that my midpoints are wrong that's I am not getting the same mean and variance as everybody else

.

Yes mean will be calculated that way.

Check the Q3 (iii) for this paper. Less than was used here too.

I found this one easier. Although its better to wait for the actual thresholds but it surely will be 35-40 range.

BTW For the histograms, did you ensure the line did not touch exactly the line for the boundaries for the right? As in <12 so being at like 11.95

The boundaries are similar. The only difference is the decimals so guess the same method will be applied.

The boundaries are similar. The only difference is the decimals so guess the same method will be applied.

the question stated repetition is allowed and less than and the numbers shkuld be less than 1000 and multiples of 5. so the last digit has onr possibility. and there were 7 no.s in total. since repetitions are allowed and it was a three digit no. 7x7 =49

Mohammad Farzanullah said:

u r forgetting that there could be 2 digits or single digit numbers also
so the answer is 57!

I wanted to ask for the histogram myself. I used 9.95, 10.45 and so on. Q4 in this paper is a similar one and exact 20, 30 are used according to the mark scheme. Both are correct perhaps.

It's hard to see that when it is computerized. I believe we are right! If not, it's one mark maximum, but we're probably right. I remember reading it in my textbook like 10 minutes before the exam and my maths teacher said:
"Yes. Just make sure the difference is only very slight."

Q1 [4 marks] (iii) 200 samples are taken. Find the probability between 163 and 173 inclusive np=160 npq=32 Less than 173.5: 0.9915 Less than 162.5: 0.6707 Between them= 0.321

Ans: (i) 0.653 (ii) 0.561 (iii) 0.321

Feel free to post any queries or mistakes

by applying continuity correction we will get P( X < = 173.5) - P( X <= 163.5) as <= 173 can be written as <174 and <= 163 can be written as <164

I also didn't drew less than 10.0 for a zero. Started with 10-10.5. Not sure if a mark will be deducted for that too.

Last Question, Last Part : The required probability was P( 163 <= X <= 173) So in other manner it can be written as P(X < = 173) - P( X< = 163) [just visualize the normal graph]

by applying continuity correction we will get P( X < = 173.5) - P( X <= 163.5) as <= 173 can be written as <174 and <= 163 can be written as <164

I did exactly the same thing!!! High-five brooo. Team NotTooSureAboutHistogramBoundaries will take over Xtremepapers! lol
Anwyays, 45/50 is what I'm expecting. Maybe 44, maybe 46. But around that!

I also didn't drew less than 10.0 for a zero. Started with 10-10.5. Not sure if a mark will be deducted for that too.

So overall not an A* and maybe not even an A. You're better off than me man!

no thats okay. you wont loose a mark dude. Well I am getting 42 max in S1, 67 max in P1, 58 max in P3 and 40 max in M1.

So overall not an A* and maybe not even an A. You're better off than me man!

no thats okay. you wont loose a mark dude. Well I am getting 42 max in S1, 67 max in P1, 58 max in P3 and 40 max in M1.

So overall not an A* and maybe not even an A. You're better off than me man!

no thats okay. you wont loose a mark dude. Well I am getting 42 max in S1, 67 max in P1, 58 max in P3 and 40 max in M1.

So overall not an A* and maybe not even an A. You're better off than me man!

Hey! The boundaries for A* have gone as low 205 or even less if I'm not mistaken and this year P1 was hard and P3 and M1 were harder from what I've heard. So you surely have a chance!
I still have to give P3 and M1 next year and even if I get 115 this year next year I'll need around 110 total for an A* and P3 is really hard.

A* has jumped upto 223 as well, so who knows? I'd want to commit suicide if I get a B. I cant handle it.
Good luck for P3 nd M1 next year

Hello guyz as its been 24 hours lets discusss...... so i hve given variant 62 for statistics...!!! first q was it 0.813 i guess

wat was de standard deviation in the last q first part??

What else did you get wrong?

I made the silliest mistake on the permutations question for 5p2. I multiplied it by 5 instead of 3 (so many 5s made me write it even though I knew I had to multiply by 3, didn't even notice). Also taking off marks which I have not identified as error,s, you know, we all have those small mistakes we think are correct and are only worth 1 mark, such as rounding to wrong amount of s.f.'s, so I'm deducting marks off myself to get in the worst case. If that makes any sense lol.

Q1 [4 marks] n=19 probability of success=0.12, find fewer than 4 so apply binomial with 0,1,2 and 3

Ans: 0.813

Q2 [4 marks] Group A was of 7, Group B was of 2, Group C was of 2. Choose 5 and at least 1 from each group. Possibilities: 3 from A and 1 from B and 1 from C. 7C3*2C1*2C1=140 2 from A and 2 from B and 1 from C. 7C2*2C2*2C1=42 2 from A and 1 from B and 2 from C. 7C2*2C1*2C2-42 1 from A and 2 from B and 2 from C. 7C1*2C2*2C2=7 Add them.

Ans: 231

Q3 (i) [3 marks] (ii) [2 marks] Roger had a probability of winning first set=0.6 Roger had a probability of winning second set if he had won the first set=0.7 Roger had a probability of winning second set if he lost the first set=0.25 Match would end when any player wins 2 sets and match cannot be drawn and Andy was the other player. (i) Probability that match would end in 2 sets. If Roger wins: 0.6*0.7=0.42 If Andy wins: 0.4*0.75=0.3 Add them. (ii) If match ends in 2 sets find the probability that match Andy won the match 0.3/0.72

Ans: (i) 0.72 (ii) 0.417

Q4 (i) [3 marks] (ii) [4 marks] (iii) [2 marks] Coin A is thrown twice and Coin B once Coin A has probability of head= 2/3 Coin B has probability of head= 1/4 (i) Show that the probability of getting 1 head is 13/36 If Coin A gives 1 head and Coin B gives none: 2*2/3*1/3*3/4=12/36 (Multiplied by 2 as both throws on Coin A can give a head) If Coin A gives 0 heads and Coin B gives 1 head: 1/3*1/3*1/4=1/36 Add them and total= 13/36 (ii) Draw a probability distribution table of numbers of heads obtained 0 when no heads obtained: 1/3*1/3*3/4=3/36 2 when Coin A gives 2 heads: 2/3*2/3*3/4=12/36 Both Coin A and Coin B give one head each: 2*1/3*2/3*1/4=4/36 (Multiplied by 2 as both throws on Coin A can give a head) Add them. 3 when all heads: 2/3*2/3:1/4=4/36 (iii) Find the expected value 1*13/36 + 2*16/36 +3*4/36=19/12

Ans : (ii) 0: 3/36 1:13/36 2: 16/36 3:4/36 (iii) 1.58

Q5 (i) [2 marks] (ii) [3 marks] (iii) [3 marks] 1345789 is a number (i) Find the number of arrangements when odd numbers together and digits not repeated 5!*3! =720 (ii) Arrangements between 3000 and 5000 and even numbers and digits not repeated When starting with 3: Last digit can be 4 or 8 so the middle two digits are 5P2=5P2*2!=40 When starting with 4: Last digit is 4 so middle two digits are 5P2=20 Total arrangements = 60 (iii) Multiples of 5 less than 1000 and digits ARE REPEATED Last digit has to be 5 and can be 3-digit, 2-digit or 1-digit number When 3 digit number: 7*7*1=49 When 2 digit number: 7*1=7 When 1-digit number: 1 Total =57

Ans: (i) 720 (ii) 60 (iii) 57

Q6 (i) [1 mark] (ii) [4 marks] (iii) [4 marks] Total athletes=57 <10 were 0 <10.5 were 4 <11.0 were 10 <12.0 were 40 <12.5 were 49 <13.5 were 57 (i) Find in the interval 10.5-11.0 =6 (ii) Draw a histogram Frequencies: 4,6,30,9,8 Frequency densities: 8,12,30,18,8 (iii) Calculate mean and variance Mid-points: 10.2, 10.7, 11.45, 12.2, 12.95 Mean=11.6 Variance=0.547

Ans: (i) 6 (iii) 11.6, 0.547

Q7 (i) [3 marks] (ii) [2 marks] (iii) [6 marks] Rafa studies at mean rate 1.9. The probability of studying more than 1.35 is 0.8 (i) Find Standard Deviation Z=0.842 so S.D. =0.653 (ii) Find the probability that studies less than 2.0 (2.0-1.9)/0.653=0.1531 P=0.561 (iii) 200 samples are taken. Find the probability between 163 and 173 inclusive np=160 npq=32 Less than 173.5: 0.9915 Less than 162.5: 0.6707 Between them= 0.321

Ans: (i) 0.653 (ii) 0.561 (iii) 0.321

Feel free to post any queries or mistakes

Heyyy dude i used the same midpoints At least someone out there who thinks alike dont know whether its correct or not but we can always pray

My prediction is that threshold for A* would be around 210 this time!

I gave all P1, P3, S1 and M1.
P1 and S1 were extremely easy while M1 and P3 were moderate with some tricky parts.

if i accidently added both the phi values and subtracted them from one how many marks will i lose

p1 65 p3 55 p4 35 p6 45 i think this will be the gt

p1 65 p3 55 p4 35 p6 45 i think this will be the gt

Q3(II) is wrong. The answer is 0.583
The conditional probability will be (0.6*0.7)/0.72 = 0.583

ATM I'm just cramming ICT theory lol. Teacher left in January and we'd only covered like half the practical exam theory and nothing else. Had to self-teach practical techniques and now learning the whole syllabus for the theory paper in 3 days. Longest hours of my life. 80 pages to memorize and learn in 3 days!! Already done 40, but my brain is exhausted! Off xtremepapers for a while. I will try to help on explaining answer solutions when I come back. Just reply to this message with your questions and I'll try explaining answers!

what will be the threshold for p32

I have a 73% in AS overall.. A in p1 D in Mechanics So average 73 it is B

After reading Kamihus solutions i feel i am gonna get 20 in Statsistics

should be happy or no.. I mean what are my grade assumptions by you all, last times AS predictions came out to be quite true!

Guys for that combination part I got all the combinations all the 4.. Fot number two, lekin I multiplied by 11C4 or something :'( Please dony ask me why but tell me how muh will I score out of 4?

Kamihus thementor

Guys I found p12 very difficult and made silly mistakes in p62 I might get about 40 but know I am hearing from people that both papers were very easy and that the gt will be higher than expected what do you think I am worried

Guys I found p12 very difficult and made silly mistakes in p62 I might get about 40 but know I am hearing from people that both papers were very easy and that the gt will be higher than expected what do you think I am worried

wrong midpoints, sorry mate. asked my teacher. in the question it was e.g <10 to <10.5. we take 10 to 10.4 and take the midpoint i.e 10.2.

If my total is about 97 do you think I can still manage an overall A

Hey can you help me too.. I dont seem to be be understanding the GT system..
So if I have a 45 in p3 and 20 in s1 does it make a B after having a 71% in my AS?
pLEASE Help me

truly speaking....with these marks u arr getting 145 marks but for B 150+ are needed

hahahahaha, how can you people be so silly man. It was continous data. Hence that is no way to take mid points. You will find out when the Ms is published

check previous mark schemes, same type of questions have come before also.

Guys i m expecting an overall 100 marks for both Maths P1 and S1, with about 60-65 in P1 and 35-40 in S1 (Inshaa Allah). Can i expect an A with these marks?

Guys i m expecting an overall 100 marks for both Maths P1 and S1, with about 60-65 in P1 and 35-40 in S1 (Inshaa Allah). Can i expect an A with these marks?

Ohh coooool! If all goes as i am expecting and i secure an A (IA) just on the edge, will it be tough to keep up with an A* in A2?

A* is around 215 marks for an average. So you'll need about 115 in the next year, so little more than A* in A2 is what you'll need.

I think i got it.. 125 for AS and 125 for A2.. A total of 250 marks, and i will be needing around 215, about 115 marks more, right?

I think i got it.. 125 for AS and 125 for A2.. A total of 250 marks, and i will be needing around 215, about 115 marks more, right?

Yeah, A* can be in the range 200-230, but for an average you can say about 215 marks in total.

Hahaha if it climbs up to 225-230, then there is absolutely no chance Lets just hope it stays low next year.. Anyways, thanks for clearing up my mind buddy

It won't be 230 as for that all four papers have to be super-easy, but this year P1 was already difficult and S1 was easy to medium. So this year thresholds have to be taken into account.

PLZZZZ answer me guys for question 5 part ii) it should be as the numbers that start by 4 =6*5*4=120 and the numbers that start by 3 = 6*5*4=120and by adding them its 24o

AM I RIGHT????

PLZZZZ answer me guys for question 5 part ii) it should be as the numbers that start by 4 =6*5*4=120 and the numbers that start by 3 = 6*5*4=120and by adding them its 24o

AM I RIGHT????

It has to be even so it will be 5*4*2 when starting with 3 as last can either be 4 or 8. It will be 5*4 as 8 will be at the end when starting with 4.

the 3 digit numbers ending in 5 = 7*7*1= 49???