What is the probability of getting 3 heads and 2 tails from 5 consecutive toss'?

[Lazerous]

I don't understand why we have to use permutation for this problem? Why is it that order matters?

All they want is 3 heads and 2 tails; they never mentioned that they want one before the other so why is it that we use permutation?

The answer is: 5/16

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[211183]

I don't understand why we have to use permutation for this problem? Why is it that order matters?

All they want is 3 heads and 2 tails; they never mentioned that they want one before the other so why is it that we use permutation?

The answer is: 5/16

Whe we say order doesn't matter... that means you have to think of ALL THE POSSIBLE orders...
when order matters its easier... because they TELL YOU waht the order is...
in this cas..

you want 3 head and 2 tails... so order does NOT matter...
so you can get

HHHTT
HHTHT
HTHHT
etc....

 

[Lazerous]

Whe we say order doesn't matter... that means you have to think of ALL THE POSSIBLE orders...
when order matters its easier... because they TELL YOU waht the order is...
in this cas..

you want 3 head and 2 tails... so order does NOT matter...
so you can get

HHHTT
HHTHT
HTHHT
etc....

That's what I am saying! I got this problem from achiever and the solution said you have to use Permutation which means order matters, right?

So what I'm not getting is why they used permutation, because I cerainly odn't get the same answer if I use nCr...

 

[doc3232]

Ok, I see what you are saying but..
Which is more likely 5 heads and 0 tails or 3 heads and 2 tails...Think about that. The 3 heads and 2 tails are much more likely because there are more ways that it can happen. The 5 heads would be heads then heads then heads...etc

 

[211183]

I don't understand why we have to use permutation for this problem? Why is it that order matters?

All they want is 3 heads and 2 tails; they never mentioned that they want one before the other so why is it that we use permutation?

The answer is: 5/16

How I would do it:

# of possible outcomes = (1/2)^6 = 1/36

Possible ways to reach outcome = 5! / (2! * 3!) = 10


so you get 10/ 32. thats the quick way. if you don't get it maybe try doing this a few times...

list possible outcomes like we did before...

HHHTT (1/2) (1/2).... etc 6 times = 1/32
HHTTH
HTTHH
TTHHH

THHHT
THHTH
THTHH
HTHHT
HHTHT

HTHTH

theres 10....



add it up... = 10/32

 

[Lazerous]

How I would do it:



Possible ways to reach outcome = 5! / (2! * 3!) = 10

See, right over here. You are using nPr, right?

Why? It doesn't matter if we get HHHTT first of if we get THHHT, just as long as we get 3 heads and two tails. Thus, why not use the combination formula, nCr?

 

[Sea of ASH]

I don't understand why we have to use permutation for this problem? Why is it that order matters?

All they want is 3 heads and 2 tails; they never mentioned that they want one before the other so why is it that we use permutation?

The answer is: 5/16

yaaaa dood i would always get stuck on these. i think exactly like that, but then i would realize if we do that, what are the odds of gettin 3heads and 2 tails in no order???? wouldnt be 100%????? cuz its only a two sided coin and odds of gettin heads or tails is equal. so if we toss a coin 5 times, it will be heads or tails 100% of the time. thats why our thought process is wrong.

now for the answer, still tryina get it myself man.

 

[211183]

The first part of the question is asking... How many ways can we get 3 heads and 2 tails??? how many COMBINATIONS of 3 heads and 2 tails are there....we know because we jsut figured it out... its 10. So we have 10 different ways of getting this possible COMBINATION. nCr

the second part of the question is asking, ok you can get 3 heads and 2 tails 10 different ways, but how many total possible ways are there in the first place?

that is 2^5.
its 2 possibilities up to the 5 tries you are taking. how do you know this? if you can't remember that its 2^5...
think of probabilty of getting all heads

HHHHH = (1/2)*(1/2)... etc.. .(1/2) ^5 = 1/32
you have a 1/32 chance of getting all heads... out of 32 possible outcomes.... which is 2*2*2*2*2

Probability = desired outcome / possible outcomes

you get 10/36

it gets compciated because you have to know that you can't answer it in just one step.

 

[Lazerous]

The first part of the question is asking... How many ways can we get 3 heads and 2 tails??? how many COMBINATIONS of 3 heads and 2 tails are there....we know because we jsut figured it out... its 10. So we have 10 different ways of getting this possible COMBINATION. nCr

the second part of the question is asking, ok you can get 3 heads and 2 tails 10 different ways, but how many total possible ways are there in the first place?

that is 2^5.
its 2 possibilities up to the 5 tries you are taking. how do you know this? if you can't remember that its 2^5...
think of probabilty of getting all heads

HHHHH = (1/2)*(1/2)... etc.. .(1/2) ^5 = 1/32
you have a 1/32 chance of getting all heads... out of 32 possible outcomes.... which is 2*2*2*2*2

Probability = desired outcome / possible outcomes

you get 10/36

it gets compciated because you have to know that you can't answer it in just one step.

See, the thing is I am understanding EVERYTHING you are saying. I know all about the multiplication rule and all of that jazz. BUT:

Combination (where order does not matter, as I thought was the case for this question): nCr= n!/[r!(n-r)!]

Permutation (where order DOES matter, and the explanation to the solution says we use permutation): nPr = n!/(n-r)!

So, that means the explanation is wrong since we have to use nCr to get 10 (and after that we divide by 32 according to the multiplication rule).

 

[211183]

maybe they are doing it opposite


nPr =

5!/3! = 20

20/32 = 10/16

thats all the ways to not get head 3 times.
so if you want to get 3 heads...

16/16- 10/16 = 5/16

u can do it either way

 

[Lazerous]

OK, how about this question:

The ratio of boys to girls in a school of 180 students is 5:4. How many new boys should join the school so that the ration becomes 3:2?

 

[211183]

OK, how about this question:

The ratio of boys to girls in a school of 180 students is 5:4. How many new boys should join the school so that the ration becomes 3:2?

Boys:Girls:Total

5 : 4 : 9

X : Y : 180





do cross multiplication

you get 100 boys and 80 girls at the ratio of 5:4


Boys : Girls : Total

3 : 2 : 5

X : 80 : X


cross multiply again

Boys = 120


You get

Boys = 120
Girls = 80

simplify 120/80 = 60/40 = 6/4 = 3 : 2


hopefully the form i used made it easier to see

 

[hoyas19]

OK, how about this question:

The ratio of boys to girls in a school of 180 students is 5:4. How many new boys should join the school so that the ration becomes 3:2?

what is the answer to this?

 

[Lazerous]

what Is The Answer To This?

20

 

[Lazerous]

Boys:Girls:Total

5 : 4 : 9

X : Y : 180





do cross multiplication

you get 100 boys and 80 girls at the ratio of 5:4


Boys : Girls : Total

3 : 2 : 5

X : 80 : X


cross multiply again

Boys = 120


You get

Boys = 120
Girls = 80

simplify 120/80 = 60/40 = 6/4 = 3 : 2


hopefully the form i used made it easier to see

Thanks a lot! That really cleared things up!

 

[Streetwolf]

maybe they are doing it opposite


nPr =

5!/3! = 20

20/32 = 10/16

thats all the ways to not get head 3 times.
so if you want to get 3 heads...

16/16- 10/16 = 5/16

u can do it either way

No. That's just wrong.

The problem is a combination. The book made a mistake and wrote permutation.

 

[211183]

thats what i thought. only explan for doing it that way i could think of. thanks for clarifying cuz that had me confused and questioning my method as well.

 

[Streetwolf]

Now that I'm looking at it closer, 16/16 - 10/16 = 6/16... not 5/16.

 

[211183]

good thing i can do mental math...

 

[211183]

and you agree with the ohter way i did it though right?

 

[Streetwolf]

Not 100%. You got lucky because the prob of a head or a tail is 1/2. Both numerators are 1. You figured out (nCr) and the denominator. You can't forget the numerator. In this case it doesn't matter since the numerators are all 1s.

But if I had a problem with a die where I wanted two 4s in 3 rolls, you would need to do (1/6)(1/6)(5/6). With your logic, you'd leave out that 5.

 

[doc3232]

Not 100%. You got lucky because the prob of a head or a tail is 1/2. Both numerators are 1. You figured out (nCr) and the denominator. You can't forget the numerator. In this case it doesn't matter since the numerators are all 1s.

But if I had a problem with a die where I wanted two 4s in 3 rolls, you would need to do (1/6)(1/6)(5/6). With your logic, you'd leave out that 5.

I hate using formulas: So what is an easy way to figure out what number needs to be multiplied by [(1/6)(1/6)(5/6)] to find the correct answer?

Thanks

 

[211183]

Not 100%. You got lucky because the prob of a head or a tail is 1/2. Both numerators are 1. You figured out (nCr) and the denominator. You can't forget the numerator. In this case it doesn't matter since the numerators are all 1s.

But if I had a problem with a die where I wanted two 4s in 3 rolls, you would need to do (1/6)(1/6)(5/6). With your logic, you'd leave out that 5.

I didn't get lucky, I knew that. And I know how to do the problem you suggested as well.

 

[Streetwolf]

I didn't get lucky, I knew that. And I know how to do the problem you suggested as well.

Well I was basing that off of a post you made above where all you did was compute nCr and the denominator.

Doc, the formula is (nCr) * p^(r) * q^(n-r)

n = # of total objects, events, etc
r = # of successes, desired objects in a group, etc
p = probability of a success
q = probability of a failure (it's just 1 - p)