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Q13 (1 Viewer)

aaron_syd

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Re: Guys does anyone have the answers for the probability for the hsc. Here are the Q

^i got the same as OP
 

scarvesss

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Re: Guys does anyone have the answers for the probability for the hsc. Here are the Q

confirmed OP
 

sghguos

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Re: Guys does anyone have the answers for the probability for the hsc. Here are the Q

thx guys confirmed, hope you got it right
 

OH NOE zzz

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Re: Guys does anyone have the answers for the probability for the hsc. Here are the Q

I got the same answers as you too :D
 

chrisv92

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damn for angle abc i put 33 degrees and the minutes following! will i lose a mark?
 

chrisv92

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That's sucks, I would be inclined to think not though. I obviously showed all working and showed I could find the angle correctly, I shouldnt get marked down for having a more specific answer
 

Joey7

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That's sucks, I would be inclined to think not though. I obviously showed all working and showed I could find the angle correctly, I shouldnt get marked down for having a more specific answer
Technically you didn't do everything the question asked, so I wouldn't say it's unreasonable for losing a mark there.
 

humayoun

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Re: Guys does anyone have the answers for the probability for the hsc. Here are the Q

Q 13 C)

Two buckets each contain red marbles and white marbles. Bucket A contains 3 red and 2 white marbles. Bucket B contains 3 red and 4 white marbles.

Chris randonly chooses one marble from each bucket

i) what is the probablility that both marbles are red

ii) what is the probability that at least one of the marbles are red

iii) what is the probability that both marbles are the same colour




My answers:

i) 9/35

ii) 26/35

iii) 9/35 +8/35 = 17/35

Hi, here is the solution:

Bucket A: (3 Red, 2 White)
Bucket B: (3 Red, 4 White)

Let:
RA = A red marble from bucket A
WA = A white marble from bucket A
RB = A red marble from bucket B
WB = A white marble from bucket B

(i) Chris chooses a marble from each bucket.

P(both are Red) = P(RA AND RB)

= P(RA) AND P(RB)

= P(RA) x P(RB)

= 3/5 x 3/7

= 9/35

(ii)

P(at least one is Red) = 1 - P(none of the two are red)

= 1 - P(both are White)

= 1- P(WA AND WB)

= 1 - (2/5 x 4/7) = 1 - 8/35

= 27/35

ALTERNATIVELY:

P(at least one is Red)

= P(RA AND WB) OR P(WA AND RB) OR P(RA AND RB)

= (3/5 x 4/7) + (2/5 x 3/7) + (3/5 x 3/7)

= (12 + 6 + 9) / 35

= 27/35

(iii)

P(both are same colour)

= P(both are Red) OR P(both are White)

= P(RA AND RB) + P(WA AND WB)

= (3/5 x 3/7) + (2/5 x 4/7)

= 9/35 + 8/35

= 17/35
 

freeeeee

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Re: Guys does anyone have the answers for the probability for the hsc. Here are the Q

Hi, here is the solution:

Bucket A: (3 Red, 2 White)
Bucket B: (3 Red, 4 White)

Let:
RA = A red marble from bucket A
WA = A white marble from bucket A
RB = A red marble from bucket B
WB = A white marble from bucket B

(i) Chris chooses a marble from each bucket.

P(both are Red) = P(RA AND RB)

= P(RA) AND P(RB)

= P(RA) x P(RB)

= 3/5 x 3/7

= 9/35

(ii)

P(at least one is Red) = 1 - P(none of the two are red)

= 1 - P(both are White)

= 1- P(WA AND WB)

= 1 - (2/5 x 4/7) = 1 - 8/35

= 27/35

ALTERNATIVELY:

P(at least one is Red)

= P(RA AND WB) OR P(WA AND RB) OR P(RA AND RB)

= (3/5 x 4/7) + (2/5 x 3/7) + (3/5 x 3/7)

= (12 + 6 + 9) / 35

= 27/35

(iii)

P(both are same colour)

= P(both are Red) OR P(both are White)

= P(RA AND RB) + P(WA AND WB)

= (3/5 x 3/7) + (2/5 x 4/7)

= 9/35 + 8/35

= 17/35
i agree with the answers sghguos got.

ii) should be 26/35 not 27/35, its a 1 mark question should be an indicator that it is just a simple subtraction
 

humayoun

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Re: Guys does anyone have the answers for the probability for the hsc. Here are the Q

i agree with the answers sghguos got.

ii) should be 26/35 not 27/35, its a 1 mark question should be an indicator that it is just a simple subtraction
How do you get 26/35? Please show your working.
 

EinstenICEBERG

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oh shit.....
l got 6/5 and 7/5 for the coordinates of N
damn...
l lost 3 marks
so that's 14 definite marks gone. l don't think l can get band 5 now :(
ye.. u didnt put the 1 on the other side to make it negative 1, just u get.. 6/5 and 7/5.. like me ..
 

freeeeee

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Re: Guys does anyone have the answers for the probability for the hsc. Here are the Q

How do you get 26/35? Please show your working.
P(At least one marble is white) = 1 - P(All Red)
which is then 1 - 9/35 = 26/35
 

Keelan134

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For 13 B(ii) I broke it up into 3 sections. I got the area for 0-3 of the concave up eqn
I then got the area from 0-4 of the concave down section, but then minused the area of the concave up eqn from 3-4

That worked for me logically in my head, anybody else think thats right or did the same?
 

humayoun

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Re: Guys does anyone have the answers for the probability for the hsc. Here are the Q

P(At least one marble is white) = 1 - P(All Red)
which is then 1 - 9/35 = 26/35
I guess it's sghguos's fault as they posted the (ii) of this question wrong!! (See below in bold)

Originally Posted by sghguos
Q 13 C)

Two buckets each contain red marbles and white marbles. Bucket A contains 3 red and 2 white marbles. Bucket B contains 3 red and 4 white marbles.

Chris randonly chooses one marble from each bucket

i) what is the probablility that both marbles are red

ii) what is the probability that at least one of the marbles are red

iii) what is the probability that both marbles are the same colour[\QUOTE]
I answered the question as posted by sghguos.

But for the question as asked in HSC Mathematics 2012, you are correct, the answer is 26/35!!
 

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