Author Topic: Exam 2 Suggested Solutions Finished (Read 27432 times) Tweet Share

/0 · « **Reply #75 on:** November 10, 2008, 05:25:21 pm »

Quote from: flubbles 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
$Pr=\frac{12}{56}=\frac{3}{14}\simeq0.214$

was this on CAS?

danieltennis · « **Reply #76 on:** November 10, 2008, 05:26:28 pm »

Quote from: DivideBy0 on November 10, 2008, 05:25:21 pm

Quote from: flubbles 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?
WTF. I dont remember this.

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
$Pr=\frac{12}{56}=\frac{3}{14}\simeq0.214$

was this on CAS?

WTF. I dont remember this.

Blizz · « **Reply #77 on:** November 10, 2008, 05:31:35 pm »

Quote from: Synesthetic 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

i thought i did, ..
then again i dont remember mayb i didnt get 2pi...

ahhh fuk it lol

lost like 6-7 in multiple......

azn_kiwi91 · « **Reply #78 on:** November 10, 2008, 05:36:49 pm »

Can someone explain to me the independent dice multi choice question? I don't get it.

sandhya · « **Reply #79 on:** November 10, 2008, 05:38:17 pm »

hey for the normal, if you drew it so that you could see the intercepts with it not extending over R, would you lose a mark?

Collin Li · « **Reply #80 on:** November 10, 2008, 05:38:53 pm »

Quote from: azn_kiwi91 on November 10, 2008, 05:36:49 pm

Can someone explain to me the independent dice multi choice question? I don't get it.

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

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

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

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

azn_kiwi91 · « **Reply #81 on:** November 10, 2008, 05:40:23 pm »

Thanks, luckily it was the only MC i didnt get

sandhya · « **Reply #82 on:** November 10, 2008, 05:41:37 pm »

what study score would you be expecting with approx 78/80, and 40/40?

Synesthetic · « **Reply #83 on:** November 10, 2008, 05:42:18 pm »

Quote from: sandhya on November 10, 2008, 05:38:17 pm

hey for the normal, if you drew it so that you could see the intercepts with it not extending over R, would you lose a mark?

It should be okay provided you drew arrow-heads at each end of the curve.

BiG DaN · « **Reply #84 on:** November 10, 2008, 05:42:30 pm »

i got 16D
and i did flip it

vcestudy123 · « **Reply #85 on:** November 10, 2008, 05:44:41 pm »

Quote from: DivideBy0 on November 10, 2008, 05:25:21 pm

Quote from: flubbles 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
$Pr=\frac{12}{56}=\frac{3}{14}\simeq0.214$

was this on CAS?

s!!

sure hope it wasnt coz i dont remember it
anyone have a CAS exam and solutions?? i really want to know what i got!!
hhaahaha
thanks heap

Synesthetic · « **Reply #86 on:** November 10, 2008, 05:45:41 pm »

Quote from: BiG DaN on November 10, 2008, 05:42:30 pm

i got 16D
and i did flip it

It's C

dV/dh = pi*r^2 => dh/dV = 1/[pi*r^2] = 1/[4pi] (r=2)
dV/dt = 2 (given)

dh/dt = dh/dV * dV/dt = 2/[4pi] = 1/[2pi] [C]

Blizz · « **Reply #87 on:** November 10, 2008, 05:49:43 pm »

Quote from: Synesthetic on November 10, 2008, 05:45:41 pm

Quote from: BiG DaN on November 10, 2008, 05:42:30 pm
i got 16D
and i did flip it
It's C

dV/dh = pi*r^2 => dh/dV = 1/[pi*r^2] = 1/[4pi] (r=2)
dV/dt = 2 (given)

dh/dt = dh/dV * dV/dt = 2/[4pi] = 1/[2pi] [C]

ah yes i think i found my problem
i subbed in r=2 into v=pi*r^2*h first up...instead of doing it later

Synesthetic · « **Reply #88 on:** November 10, 2008, 05:51:23 pm »

Quote from: Blizz on November 10, 2008, 05:49:43 pm

Quote from: Synesthetic on November 10, 2008, 05:45:41 pm
Quote from: BiG DaN on November 10, 2008, 05:42:30 pm
i got 16D
and i did flip it
It's C

dV/dh = pi*r^2 => dh/dV = 1/[pi*r^2] = 1/[4pi] (r=2)
dV/dt = 2 (given)

dh/dt = dh/dV * dV/dt = 2/[4pi] = 1/[2pi] [C]

ah yes i think i found my problem
i subbed in r=2 into v=pi*r^2*h first up...instead of doing it later

That doesn't make a difference

BiG DaN · « **Reply #89 on:** November 10, 2008, 05:52:35 pm »

hmm
i think i flipped dv/dt

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