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How was 9709_w15_qp_62?

Difficulty level.


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Hell yeah its 480. 5! For 5remaining seats and 2!*2! For j and s arranged between 3seats while being together.
Also the 30 is obvious 6C4*4C4*2

But what about positive/even? Any experts I did it wrong and got 0.5
I am glad that i did that permutation one right!
 
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1. n = 22
2. 9C6 / 16C6 = 3/286
3.(i) 1/4
(ii) 27/256 or 0.105 or (0.25)(0.75)^3
(iii) 3/32 or 0.09375
4.(i) 30
(ii) 480
5. IQR = probably 19? Forgot my answer
think my mean was 103 kg-ish tho
6. A {1, 2, 3, 3}
B {-3, -2, -1, 1}


A
| 1 2 3 3
_____________
-3 | -2 -1 0 0
B -2 | -1 0 1 1
-1 | 0 1 2 2
1 | 2 3 4 4

p(X = -2) = 1/16
p(X = -1) = 1/8
p(X = 0) = 1/4
p(X = 1) = 3/16
p(X = 2) = 3/16
p(X = 3) = 1/16
p(X = 4) = 1/8

Var(X) = 23/8
p(even | positive integer) = 5/9
7. Think it was 316 days, 0.840 and 0.933
what were the total marks for Var part?
 
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Any one remembers the value of m in question 7?
Mine was something 743019.25 sth Its just random numbers :p It was a big number from 7 right?
 
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Yo any idea where you messed up? I got final ans 1/2 too but i remember doing the table fine(checked but adding everything they became 1. But i too messed up the varience my mean wasnt exactly 1
I accidentally thought there were 20 values, instead of 16. A simple error which will cost me quite a few marks.
 
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I accidentally thought there were 20 values, instead of 16. Aine error which will cost me quite a few marks.
Noooooo even if you get 20 or 16, the value for the probability should be same cause denominator gets cancelled in the division.. You must have done another mistake to get1/2 as answer like me
 
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Noooooo even if you get 20 or 16, the value for the probability should be same cause denominator gets cancelled in the division.. You must have done another mistake to get1/2 as answer like me
Not true. What was your original denominator?
I counted every value in the table, so my numerator changed as well.
 
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Yo guys what was the answer to question 6 part iv) for finding the probability of the even numbers but positive integers?I first wrote P(2,4)= 1\4 but then i corrected it to P(0,2,4)=9/16
 
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Guys.. For 4a) I got 1440.. 2*2*6c2*4! the reason is.. the first 2 because the two people can interchange, the second 2 so that they can switch cars, 6C2 for the Combinations of people who sit in the car with the two people and 4! for the arrangement. Its not 4C4, That give you only 1 combination.

4b) i got 480.. 2*2*5!
 
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