The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Edexcel S2 - binomial is pissing me off... (question)

wow apparently i can't do the first question on the s2 paper.. no surprise though, binomials are confusing as ****.

question:


A fair coin is tossed 4 times. Find the probability that

(a) an equal number of head and tails occur

ok, usually I would have done 4C2 (0.5)^2 (0.5)^2 (which is the correct answer). but my teacher introduced me to this little formula yesterday for binomials:

n!/r!(n-r)!

so the working ends up being 4!/2!2! (0.5)^2 (0.5)^2 :confused:

they give different results.

1) what's the difference between nCr and n!/r!
2) under what circumstances to i use nCr and under what circumstances do i use n!/r!(n-r)! ?

if somebody can clear this up for me it would be much appreciated. this is a subject that really get's on my man tits. thanks in advance!
Reply 1
Avatar for Noble.
Noble.
17
Original post by ucasisannoying
wow apparently i can't do the first question on the s2 paper.. no surprise though, binomials are confusing as ****.

question:


A fair coin is tossed 4 times. Find the probability that

(a) an equal number of head and tails occur

ok, usually I would have done 4C2 (0.5)^2 (0.5)^2 (which is the correct answer). but my teacher introduced me to this little formula yesterday for binomials:

n!/r!(n-r)!

so the working ends up being 4!/2!2! (0.5)^2 (0.5)^2 :confused:

they give different results.

1) what's the difference between nCr and n!/r!
2) under what circumstances to i use nCr and under what circumstances do i use n!/r!(n-r)! ?

if somebody can clear this up for me it would be much appreciated. this is a subject that really get's on my man tits. thanks in advance!


nCr=n!r!(n−r)!_nC_r = \frac{n!}{r!(n-r)!}

So

4C2=4!2!(2)!_4C_2 = \frac{4!}{2!(2)!}

Basically, they're identical. :smile:

When you do nCr on your calculator, that's what it's doing. So it can't give different results.
Reply 2
Avatar for ucasisannoying
ucasisannoying
OP
Original post by Noble.
nCr=n!r!(n−r)!_nC_r = \frac{n!}{r!(n-r)!}

So

4C2=4!2!(2)!_4C_2 = \frac{4!}{2!(2)!}

Basically, they're identical. :smile:

When you do nCr on your calculator, that's what it's doing. So it can't give different results.
oh, turns out i do end up with the same answer lol, thanks...

p.s. i hate stats
Reply 3
Avatar for ucasisannoying
ucasisannoying
OP
another thing...

(b) all the outcomes are the same (answer is 0.125)
(c) the first tail occurs on the third throw.

any idea how to do part (c)?

(thanks again!)
Reply 4
Avatar for Noble.
Noble.
17
Original post by ucasisannoying
another thing...

(b) all the outcomes are the same (answer is 0.125)
(c) the first tail occurs on the third throw.

any idea how to do part (c)?

(thanks again!)


Just the same as before, except there now aren't combinations. The first two throws have to be heads.

So you'd have HHT...
Reply 5
Avatar for ucasisannoying
ucasisannoying
OP
Original post by Noble.
Just the same as before, except there now aren't combinations. The first two throws have to be heads.

So you'd have HHT...
iyt iyt...

so it's just 0.5^3?
Reply 6
Avatar for Noble.
Noble.
17
Original post by ucasisannoying
iyt iyt...

so it's just 0.5^3?


No. If you toss the coin four times, and the first two have to be heads, and the third tails, what can the fourth result be?
Reply 7
Avatar for ucasisannoying
ucasisannoying
OP
Original post by Noble.
No. If you toss the coin four times, and the first two have to be heads, and the third tails, what can the fourth result be?
could be either heads or tails...?
Reply 8
Avatar for Noble.
Noble.
17
Original post by ucasisannoying
could be either heads or tails...?


Yes, so the probability of the first tails appearing on the third throw will be the sum of the two probabilities HHTH and HHTT.
Reply 9
Avatar for ucasisannoying
ucasisannoying
OP
Original post by Noble.
Yes, so the probability of the first tails appearing on the third throw will be the sum of the two probabilities HHTH and HHTT.
(0.5)^4 + (0.5)^4 is the same as (0.5)^3 lol :tongue:

your method makes more sense though, i'll use that if I get another question like this. thanks for everything man
Reply 10
Avatar for Noble.
Noble.
17
Original post by ucasisannoying
(0.5)^4 + (0.5)^4 is the same as (0.5)^3 lol :tongue:

your method makes more sense though, i'll use that if I get another question like this. thanks for everything man


Ah yes.

2(0.5)4=0.5−1×0.54=0.532(0.5)^4 = 0.5^{-1} \times 0.5^4 = 0.5^3

I thought you were just doing

P(HHT)=0.5×0.5×0.5P(HHT) = 0.5 \times 0.5 \times 0.5

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you