Mathematics Marathon HSC 09 (2 Viewers)

[untouchablecuz]

i made this one up

the probability that a person has red eyes is 0.34 and the probability that a person has green hair is 0.65

i) find the probability that a person has both red eyes and green hair

ii) if a group of 5 people are chosen, find the probability that at least 3 of them have both red eyes AND green hair

 

[deviousgeek]

Skytopia : Super impossible math questions


Go Go Go.

 

[mtsmahia]

LOL tranny this question came up in my half yearlies. I freaked out when i saw it and lost 8 marks like that
Now it is the easiest question lol.
i) x+2 = x^2 - 5x + 11
.:x^2-6x+9=0
.x-3)^2=0
.:Since only one solution it is a tangent.
ii)At x=3 y=5
.:Mn=-1
.:y-5=-1(x-3)
.:y=-x +8
iii)Find the x-ordinates where they intersect then integrate -x + 8 - (x^2 - 5x + 11 ) dx.

umm just one Q...wouldnt u need to find the discriminant of x^2 -6x +9=0 and then evaluate if it is a tangent? . Because, even if there is one solution, cant that still mean that the line y=x+2 intersects the parabola?

 

[boxhunter91]

mtsmahia you can do that. But i've been told by my teacher that it is fine to prove by my method.

 

[mtsmahia]

mtsmahia you can do that. But i've been told by my teacher that it is fine to prove by my method.

k cool

 

[boxhunter91]

Lets keep this thread moving.
A total of 300 tickets are sold in a raffle which has 3 prizes. There are 100 red, 100 green and 100 blue tickets.
At the drawing of the raffle, winning tickets are NOT replaced before the next draw.
i) what is the probability that each of the 3 winning tickets is Red?
ii) What is the probability that at least one of the winning tickets is not red?
iii) What is the probability that there is one winning ticket of each colour?

 

[untouchablecuz]

i made this one up

the probability that a person has red eyes is 0.34 and the probability that a person has green hair is 0.65

i) find the probability that a person has both red eyes and green hair

ii) if a group of 5 people are chosen, find the probability that at least 3 of them have both red eyes AND green hair

noone?

 

[boxhunter91]

What happened to the 0.01?

 

[untouchablecuz]

no no

their not complementary events; the probability of one is independent of the other

 

[ForrestGump]

Lets keep this thread moving.
A total of 300 tickets are sold in a raffle which has 3 prizes. There are 100 red, 100 green and 100 blue tickets.
At the drawing of the raffle, winning tickets are NOT replaced before the next draw.
i) what is the probability that each of the 3 winning tickets is Red?
ii) What is the probability that at least one of the winning tickets is not red?
iii) What is the probability that there is one winning ticket of each colour?

Geeze you get around boxhunter91 hahah every thread I read you've posted on hahah. Okay to your question:
a) P(RRR) = P(R) x P(R) X P(R)
= 1/3 X 99/299 X 98/298
= 1617/44551 = about 3.6%
b) P(Not red) = 1 - P(RRR)
= 42934/44551 = about 96.4%
c) P(RBG) = P(RBG) + P(BGR) + (RGB) + P(BRG) + P(GBR) + P(GRB)
= 6 x (100/300 X 100/299 X 100/298)
= 0.224461852 = approx 22.4%

Tell me if thats right before i post another question

 

[boxhunter91]

Yeh its right. Good work.

 

[ForrestGump]

Booyah! okay next question:

A ball is dropped from a height of 2 metres onto a hard floor and bounces. After each bounce, the maximum height reached by the ball is 75% of the previous maximum height. Thus, after it first hits the floor, it reaches a height of 1.5 metres before falling again, and after the second bounce it reaches a height of 1.125 metres before falling again.

a) what is the maximum height reached after the 3rd bounce
b) what kind of sequence is formed by the successive maximum heights
c) what is the maximum distance travelled by the ball from the time it was dropped until it eventually comes to rest on the floor.
(altogether about 7 marks from a past paper)

 

[boxhunter91]

I remember doing this one.
i.T4=Ar^3 = 2*(3/4)^3= 0.84375m
ii. Geometric Sequence -->Limiting sum.
iii.I did this a different way to the Success One answers.
Limiting Sum = a/1-r
.: 1.5+2/1-3/4
=3.5/0.25
=14m.

 

[ForrestGump]

i made this one up

the probability that a person has red eyes is 0.34 and the probability that a person has green hair is 0.65

i) find the probability that a person has both red eyes and green hair

ii) if a group of 5 people are chosen, find the probability that at least 3 of them have both red eyes AND green hair

Okay:

i) P(RG) = P(R) x P(G)
= 34/100 X 65/100
= 221/1000 or 22.1%

ii) P( at least 3 RG) = 1 - (P(none) + P(1RG) + P(2RG)
= 1 - ((0.779)^5 + 5(22.1/100x77.9/100^4) + 10((22.1/100)^2 x (77.9/100)^3))
= 0.075320058 or 7.5%

I'm pretty sure thats right, check it for me.

 

[ForrestGump]

I remember doing this one.
i.T4=Ar^3 = 2*(3/4)^3= 0.84375m
ii. Geometric Sequence -->Limiting sum.
iii.I did this a different way to the Success One answers.
Limiting Sum = a/1-r
.: 1.5+2/1-3/4
=3.5/0.25
=14m.

You probs already checked it but yeah thats right.
okay give us another

 

[boxhunter91]

Solve:

 

[ForrestGump]

nope, can't do it. i know it uses a rule to switch them all around and get rid of the log but i can't remember it.
Does it go something like: (3x - 4) = 2^5 ?

 

[boxhunter91]

Come on Forrest!
x=loga^N
.:N=a^x

so 2^5=3x-4
.:32=3x-4
.:36=3x
.:x=12

 

[untouchablecuz]

i made this one up

the probability that a person has red eyes is 0.34 and the probability that a person has green hair is 0.65

i) find the probability that a person has both red eyes and green hair

ii) if a group of 5 people are chosen, find the probability that at least 3 of them have both red eyes AND green hair

i) P(person is RG)=(0.65*0.34)=0.221

ii) P(person is not RG)=0.779

P(at least 3 RG)=1-P(no RG, 1 RG, 2 RG)

P(no RG, 1 RG, 2 RG)=(0.779)⁵+⁵C₁(0.221)¹(0.779)⁴+⁵C₂(0.221)²(0.779)³

.'. P(at least 3 RG)=1-(0.779)⁵-⁵C₁(0.221)¹(0.779)⁴-⁵C₂(0.221)²(0.779)³

 

[untouchablecuz]

nope, can't do it. i know it uses a rule to switch them all around and get rid of the log but i can't remember it.
Does it go something like: (3x - 4) = 2^5 ?

one way to do it that doesn't require you to memorise anything

log₂(3x-4)=5

raise both sides to the power of 2

2^log₂(3x-4)=2⁵

taking the logarithm of base 2 of a number or expression is the inverse operation of raising it to the power of 2 --> hence, they cancel out

so, (3x-4)=2⁵ --> x=(2⁵+4)/3=12