[urgent] harder 3 unit help! (1 Viewer)

[AbsoluteValue]

A and B have (n+1) and n fair coins respectively and they toss their coins simultaneously.

a) Show that the probability that B gets r heads is given by:


b) Hence, find the probability that A will obtain k more heads than B, where:


c) Show that the probability that A will obtain more heads than B is:


Probability is arguably the hardest topic in 4 unit I really need the solution to this asap.

 

[AbsoluteValue]

help me!

 

[xamineO]

Well this question is simple.
The answer is 1/2 For part c)
Very easy question for a 4uniter.

 

[AbsoluteValue]

well this question is simple.
The answer is 1/2 for part c)
very easy question for a 4uniter.

lol -_- this question is VERY hard, i need help

 

[AbsoluteValue]

I honestly have no idea what to do, can a real mathematician offer some help here?

 

[xamineO]

Well a coin only has two probabilities thus the answer is 1/2
Even a student who didn't do maths know this.
Come on man.

 

[AbsoluteValue]

Well a coin only has two probabilities thus the answer is 1/2
Even a student who didn't do maths know this.
Come on man.

This is binomial probability, it involved more than only one trial...
and it says A have 1 more coin than B or whatever...

 

[AbsoluteValue]

Bump!

 

[AbsoluteValue]

Come on, anyone? I really need help with this, this question is a monster.

 

[twinklegal19]

be patient

I'm sure a BOSer will be able to help you soon

 

[xDarkSilent]

dont you use the binomial expansions of the n+1 terms and the nth terms ?

 

[seanieg89]

Well a coin only has two probabilities thus the answer is 1/2
Even a student who didn't do maths know this.
Come on man.

It isn't THAT obvious...



My solution:

a) comes straight from the definition of binomial probability.


For b), let X be the number of heads thrown by A and Y the number of heads thrown by B. Since the probability of any individual flip being heads is 1/2, the probability of A throwing k more heads than B is the same as the probability of A throwing k more heads than the number of tails B throws.

ie P(X-Y=k)=P(X-(n-Y)=k)=P(X+Y=n+k)

which is the just a way of writing "the probability of exactly n+k heads showing after 2n+1 flips".

So the answer to b), using binomial probability again, is:


For c), using the same trick as in b): P(X>Y)=P(X>(n-Y))=P(X+Y>n). But the complement of the event X+Y>n is the event "The total number of tails exceeds n", an event which is equally likely by symmetry. Hence P(X>Y)=1/2.

 

[xDarkSilent]

If you find a maths question TOO EASY TO DO - Especially a harder 3u Question.

It means that you are doing it wrong T_T /

 

[AbsoluteValue]

It isn't THAT obvious...



My solution:

a) comes straight from the definition of binomial probability.


For b), let X be the number of heads thrown by A and Y the number of heads thrown by B. Since the probability of any individual flip being heads is 1/2, the probability of A throwing k more heads than B is the same as the probability of A throwing k more heads than the number of tails B throws.

ie P(X-Y=k)=P(X-(n-Y)=k)=P(X+Y=n+k)

which is the just a way of writing "the probability of exactly n+k heads showing after 2n+1 flips".

So the answer to b), using binomial probability again, is:


For c), using the same trick as in b): P(X>Y)=P(X>(n-Y))=P(X+Y>n). But the complement of the event X+Y>n is the event "The total number of tails exceeds n", an event which is equally likely by symmetry. Hence P(X>Y)=1/2.

Thanks a lot man.

 

[seanieg89]

No worries.

 

[Come at me]

Which year?

 

[AbsoluteValue]

Which year?

HSC

 

[Come at me]

no shit, lol
which year?

HSC

 

[umm what]

2000 isn't it?

 

[AbsoluteValue]

No, this question is from Sydney Grammar trials I think... I thought u were asking which year I'm in, in that case year 12, lol my bad.