Author Topic: Exam 2 Suggested Solutions Finished (Read 27222 times) Tweet Share

trinon · « **Reply #60 on:** November 10, 2008, 04:42:18 pm »

Finished.

dekoyl · « **Reply #61 on:** November 10, 2008, 04:45:38 pm »

Thank you. Amazing how fast you do them o_O The exam finished 2 hrs 45 minutes ago (seems like a really short time to me.)

Butler · « **Reply #62 on:** November 10, 2008, 04:50:00 pm »

For question 1ciii, I think you've got the denominator wrong. Shouldn't it be the answer to the previous question?

doboman · « **Reply #63 on:** November 10, 2008, 04:50:50 pm »

He acknowledged it

Mikey123 · « **Reply #64 on:** November 10, 2008, 04:55:49 pm »

Thnaks buddy.

Butler · « **Reply #65 on:** November 10, 2008, 04:59:00 pm »

Quote from: doboman on November 10, 2008, 04:50:50 pm

He acknowledged it

Apologies. Thanks Trinon.

flubbles · « **Reply #66 on:** November 10, 2008, 04:59:33 pm »

Quote from: 1ciii

What is the probability, correct to three decimal places, that her first 4 attempts at scoring a goal are successful, given that exactly 6 of her first 8 attempts at scoring a goal in a match are successful?

Combinatorics solution:

Given exactly 6 successes == given exactly 2 failures
Our probability space has these 2 failures anywhere in the 8 attempts, so 8 choices for the first fail * 7 choices for the second fail = 56
The event we want has these 2 failures anywhere in the last 4 attempts, so 4 choices for the first fail * 3 choices for the second fail = 12
$Pr=\frac{12}{56}=\frac{3}{14}\simeq0.214$

hanhanchampion · « **Reply #67 on:** November 10, 2008, 05:00:54 pm »

Quote from: coblin on November 10, 2008, 04:22:32 pm

Quote from: hanhanchampion on November 10, 2008, 04:19:43 pm
15 is A... they are independent which means Pr(a/b)=Pr(a) I checked the book. So A must be contained or contain B. Only A is right. I think it's A{1,2,4} B{1,2} or something like that..

Yes, independent means: $\mbox{Pr}(A|B) = \mbox{Pr}(A)$ , which implies:

$\frac{\mbox{Pr}(A \cap B)}{\mbox{Pr}(B)} = \mbox{Pr}(A)$

$\mbox{Pr}(A \cap B) = \mbox{Pr}(A) \times \mbox{Pr}(B)$

The answer is E. See this post. I have modified it so the first case is E, and the last case is A:

Quote from: coblin on November 10, 2008, 03:27:07 pm
Quote from: Nelle91 on November 10, 2008, 03:24:28 pm
I got 15 B aswell but it seemed like the most logical choice, had no idea lol

You work out Q15 by verifying this:

$\mbox{Pr}(A \cap B) = \mbox{Pr}(A) \times \mbox{Pr}(B)$

For example (option E):

If $A = \{ 1,2\}$ , $B = \{ 2,4,6\}$ , then: $A \cap B = \{2 \}$

$\mbox{Pr}(A) = \frac{1}{3}$

$\mbox{Pr}(B) = \frac{1}{2}$

$\mbox{Pr}(A \cap B) = \frac{1}{6}$

Since $\mbox{Pr}(A) \times \mbox{Pr}(B) = \frac{1}{6} = \mbox{Pr}(A \cap B)$ , then events $A$ and $B$ are independent.

Another example (option A):

If $A = \{ 1,2,3\}$ , $B = \{ 1,2\}$ , then: $A \cap B = \{1,2 \}$

$\mbox{Pr}(A) = \frac{1}{2}$

$\mbox{Pr}(B) = \frac{1}{3}$

$\mbox{Pr}(A \cap B) = \frac{1}{3}$

Since $\mbox{Pr}(A) \times \mbox{Pr}(B) = \frac{1}{6} \neq \mbox{Pr}(A \cap B)$ , then events $A$ and $B$ are not independent!

ok.... i understand... but be a silly mistake..

vcestudy123 · « **Reply #68 on:** November 10, 2008, 05:06:42 pm »

Hey.. just wondering if anyone has a copy of the CAS solutions.. i know most questions were the same but there were a few differences
thanks

Synesthetic · « **Reply #69 on:** November 10, 2008, 05:08:28 pm »

Thanks trinon;

My worded answer to 2ciii (justification of integral from 1 to a) was almost identical to yours, and I think it is correct.

For 3h, I think 6 days is correct, as long as you explicitly say he takes 6 capsules in the 6 days - so you probably got the mark in the exam. If you state 5 days (like I did) they may possibly accept that too because he takes the 6 capsules over 5 days.

As others and yourself point out, 1aiii is wrong; it's 0.214 as according to flubbles.

3d: a sign diagram is probably necessary.

A final note, for the sketch on the last page, x-intercepts are not required, though including them as you did won't hurt!

Blizz · « **Reply #70 on:** November 10, 2008, 05:15:57 pm »

is question 10 really C? shouldnt it be B (brackets)

Synesthetic · « **Reply #71 on:** November 10, 2008, 05:18:23 pm »

Quote from: Blizz on November 10, 2008, 05:15:57 pm

is question 10 really C? shouldnt it be B (brackets)

10 is definitely C; a maximum point of the curve was 2, C was the only choice with ...,2] for its range.

Blizz · « **Reply #72 on:** November 10, 2008, 05:21:35 pm »

Quote from: Synesthetic on November 10, 2008, 05:18:23 pm

Quote from: Blizz on November 10, 2008, 05:15:57 pm
is question 10 really C? shouldnt it be B (brackets)

10 is definitely C; a maximum point of the curve was 2, C was the only choice with ...,2] for its range.

oh bugger

thought it was ....,2) .. not ']'

is 16 D :unsure: i thought i clearly got 2pi not 1/2pi

ahhhh

lol i got that probablity independent one right. i saw that i had not chosen E for a while so i just guessed it xD

shuie · « **Reply #73 on:** November 10, 2008, 05:22:38 pm »

FAR OUT.

Synesthetic · « **Reply #74 on:** November 10, 2008, 05:24:41 pm »

Quote from: Blizz on November 10, 2008, 05:21:35 pm

Quote from: Synesthetic on November 10, 2008, 05:18:23 pm
Quote from: Blizz on November 10, 2008, 05:15:57 pm
is question 10 really C? shouldnt it be B (brackets)

10 is definitely C; a maximum point of the curve was 2, C was the only choice with ...,2] for its range.
oh bugger

thought it was ....,2) .. not ']'

is 16 D :unsure: i thought i clearly got 2pi not 1/2pi

ahhhh

lol i got that probablity independent one right. i saw that i had not chosen E for a while so i just guessed it xD

No, 16 is C - you probably forgot to flip dV/dh to dh/dV to get dh/dt

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