AQA Maths Linear June 2011 Calculator paper (Specification A)

Reply 80

Since95

did anyone get an inkling of an answer for Q24a?

Reply 81

Member737,514

10

Original post by Since95

the question was using the cards 1-9 , 9 cards, pick two cards to make atleast 16 what is the probability (not entirely sure whether it said it was replaced, but i'm assuming it did... as it definitely didn't say otherwise)

anyway first for the question you had to work out the pairs that mad 16+

7+9=16

8+9=17

7=1/9
8=1/9
9=1/9

as a fraction for probability

so first you'll find the probability of getting one of the sums - the and rule.

1/9 x 1/9 = 1/81 - probability of 7+9

and the same for 8+9

1/9 x 1/9 = 1/81

which gave the same probability.

then you would use the or rule of getting either pairs (adding the fractions)

1/81 + 1/81 = 2/81

the final probability should have been 2/81.

please let me know if you spot any error, but i feel it is correct (challenge me :confused:

)

But wouldn't the probability only be out of 81 if there were two sets of these nine cards? The, it would be 9 x 9 = 81 combos. If you add the possibilities when withdrawing two cards simultaneously from the 9 available, then from 1-9 there are eight possibilities (1 and 2, 1 and 3, 1 and 4 etc up to and including 9) and as you go along the number 'line' consecutively the possibilities decrease by one each time, resulting in 8 [1 to 9] + 7 [2 to 9] + 6 [3 to 9] + 5 [4 to 9] + 4 [5 to 9] + 3 [6 to 9] + 2 [7 to 9] +1 [8 to 9] = 36. The cards were not replaced so surely there are 36 combinations? :smile:

Reply 82

Member737,514

10

Original post by Since95

did anyone get an inkling of an answer for Q24a?

Which question was that?

Reply 83

glurblug

Original post by Since95

the question was using the cards 1-9 , 9 cards, pick two cards to make atleast 16 what is the probability (not entirely sure whether it said it was replaced, but i'm assuming it did... as it definitely didn't say otherwise)

anyway first for the question you had to work out the pairs that mad 16+

7+9=16

8+9=17

7=1/9
8=1/9
9=1/9

as a fraction for probability

so first you'll find the probability of getting one of the sums - the and rule.

1/9 x 1/9 = 1/81 - probability of 7+9

and the same for 8+9

1/9 x 1/9 = 1/81

which gave the same probability.

then you would use the or rule of getting either pairs (adding the fractions)

1/81 + 1/81 = 2/81

the final probability should have been 2/81.

please let me know if you spot any error, but i feel it is correct (challenge me :confused:

)

Did they specify the numbers had to be different? Because if it was with replacement you'd have to count the possibility of picking a 9 out of the bag again if you got a nine in the first place (or 8 and 8), you could get 9&9, 8&8 for pairs of at least 16. But since when were two of the same number a "pair" of numbers? A "pair" to me sounds like two different numbers you hold in your hand. It's just looking very likely to me that this was without replacement.

If it was with replacement and the numbers had to be different, that would just be bizarre, pointless and would be basically the same as being without replacement (1/9 * 1/8). So either it's with replacement and you can get "pairs" of the same 9 and the same 8, or without replacement (which makes a lot more sense to me).

(edited 12 years ago)

Reply 84

bluekipper

That’s right - If the first card was replaced, then the same card could be picked a second time, so the probability would be 4/81.

However, the first card was not replaced, so the correct solution is:
(1/9 x 1/8) + (1/9 x 1/8) + (1/9 x 2/8) = 4/72 = 1/18

Reply 85

glurblug

Original post by bluekipper

That’s right - If the first card was replaced, then the same card could be picked a second time, so the probability would be 4/81.

However, the first card was not replaced, so the correct solution is:
(1/9 x 1/8) + (1/9 x 1/8) + (1/9 x 2/8) = 4/72 = 1/18

That's what I got for the answer. All questions I've seen in the past like this one have just been straightforward "without replacement" questions, with the aim of catching people out who'll forget you can get 7 then 9 as well as 9 then 7. All this "with replacement" and "taking both numbers out at exactly the same time" stuff just overcomplicates it and doesn't make sense.

(edited 12 years ago)

Reply 86

Since95

Original post by bluekipper

That’s right - If the first card was replaced, then the same card could be picked a second time, so the probability would be 4/81.

However, the first card was not replaced, so the correct solution is:
(1/9 x 1/8) + (1/9 x 1/8) + (1/9 x 2/8) = 4/72 = 1/18

My solution was wrong..... i guess?

but it all comes down to whether the card is replaced, or not.
i think yours is correct but otherwise if the card was replaced the answer would be 4/81

Reply 87

Since95

Original post by Member737,514

Which question was that?

i remeber the first part involved subsitution wher in an exuation x=-1
and by subsituting with -1 the answer should have been 5/2(i think)
all i kept on getting was 1/8 or 1/10

Reply 88

Member737,514

10

Original post by Since95

i remeber the first part involved subsitution wher in an exuation x=-1
and by subsituting with -1 the answer should have been 5/2(i think)
all i kept on getting was 1/8 or 1/10

Oh I remember now. I got 4/10 which cancelled down to 2/5. Initially I thought we had to solve it to prove that x=-1 so I spent quite a long time factorising before I realised it was a substitution question. :redface:

Did you get 2/5 in the end?

Reply 89

Mazin EL

Did any1 get that h for the triangle was like 5.3 because you needed the max area but min base or summin.
And what did people get for the last vector question where you had to prove that ACG lie on a straight line?

(edited 12 years ago)

Reply 90

boring exams

for the 24picm^ question, i made a silly mistake and i simplified 60/360 to 1/60 instead of 1/6, so i got 1440 (instead of 144) and i square rooted it and i got something like 39.7. Will i get any marks for method? :frown:

And how many marks was probability question?

Reply 91

Mazin EL

the probability question was 4 marks and what did u get for the very last question?

Reply 92

boring exams

what did everyone get for last question? I said vectors ac and ag were in ratio 1:2. is that right???

Reply 93

Since95

Original post by boring exams

for the 24picm^ question, i made a silly mistake and i simplified 60/360 to 1/60 instead of 1/6, so i got 1440 (instead of 144) and i square rooted it and i got something like 39.7. Will i get any marks for method? :frown:

And how many marks was probability question?