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OCR (Not MEI) S1 Wednesday 8th June 2016

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Reply 100
Original post by XBen1999
What did people get for the question where it said what is the p of X equaling each other and how did you do it


I thought x could be 00 or 11 or 22 so i did the prob of getting 0 squared add prob of getting 1 squared add prob of getting 2 squared
Original post by catwat99
I got 1/70 for the ABBA question??


Yeah same! It's wrong tho
Original post by Breynolds2671999
Same, just did ((1x0.6 + 17)) + ((2x0.6 + 17))...... and i think i got that.


yeah i didnt know how else to work it out so typed it all out in the calculator aha
I got 3/35
Original post by cc262626
Yeah same! It's wrong tho
Original post by cc262626
Yeah same! It's wrong tho


oh, why?
Original post by XBen1999
What did people get for the question where it said what is the p of X equaling each other and how did you do it


P(X=0)^2 + P(X=1)^2 + P(X=2)^2, found each individual probability just using the binomial formula
Original post by iamDev
I thought x could be 00 or 11 or 22 so i did the prob of getting 0 squared add prob of getting 1 squared add prob of getting 2 squared


Yeh I did that but I think I got the wrong answer how many marks will I lose?
Reply 107
S2g im only person who got 2/35 for ABBA 😂😂 what was the actual answer?
For the probability Q with x<9<20 I just did sum of probability a for 10-19 ... And I got like 0.093 what were u supposed to do


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Original post by student199919
yeah i didnt know how else to work it out so typed it all out in the calculator aha


Same i got so confused , i just wanted a nice easy working oit a regression line and we got that :biggrin:
Original post by cc262626
Yeah same! It's wrong tho


I'm pretty sure it's right? I got that but wrote it as 0.0143
Original post by duncant
1/105 probability to get ABBA? only one im not sure on

I got 2/35
Original post by Shanie_e
I'm pretty sure it's right? I got that but wrote it as 0.0143


Are you sure?! Loads at my school said was wrong
Original post by h3rmit
P(X=0)^2 + P(X=1)^2 + P(X=2)^2, found each individual probability just using the binomial formula


I thought i did it wrong, can't believe i actually got it right :biggrin:
Original post by Ella rivers1
For the probability Q with x<9<20 I just did sum of probability a for 10-19 ... And I got like 0.093 what were u supposed to do


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.8^9 - .8^19
The same thing was in a past paper, but I couldn't remember what to do with the indices so I just put .2(.8^9 + .8^10 ... + .8^19) in my calculator
Original post by cc262626
Yeah same! It's wrong tho


How?? I did (3x2x1x2)/7P4
Reply 116
If I got the formula for variance wrong, will i get error carried forward for the next question?
Yeah same ! But I ended up with wrong number lol


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Original post by Ella rivers1
For the probability Q with x<9<20 I just did sum of probability a for 10-19 ... And I got like 0.093 what were u supposed to do


Posted from TSR Mobile


I just worked out the probability for 10, 11, 12, 13, 14, 15, 16, 17, 18 and 19 and added them together.
Original post by Shanie_e
I'm pretty sure it's right? I got that but wrote it as 0.0143


my first answer was 12/35 because i did something like
(3C1 X 2C1 X 1C1 X 2C1)/7C4 and got 12/35. then i wanted to see if i got the same with another method and so i did 3/7 x 2/6 x 1/5 x 2/4 and got 1/70. so im not sure

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