No understanding of probability whatsoever.

Darketernal

New member
Joined
Mar 14, 2017
Messages
4
I must be the worst person at math in the universe, oh well here goes.

My problem is that i don't understand the explanation (i don't know how to calculate it).

The probability question is as following:

You have 6 coins, what is the probability that you throw 4 times tails.

La place definition is the amount of beneficial outcomes divided by the amount of total possible outcomes.

so far so good.

The amount of total possible outcomes = 2*2*2*2*2*2 = 64

And here is where i stumble.

The description is (6)
(4) = 15 Beneficial outcomes

And i am completely lost

what do they mean by (6)
(4) = 15

And how to calculate it? :confused:
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,945
I must be the worst person at math in the universe, oh well here goes.

My problem is that i don't understand the explanation (i don't know how to calculate it).

The probability question is as following:

You have 6 coins, what is the probability that you throw 4 times tails.

La place definition is the amount of beneficial outcomes divided by the amount of total possible outcomes.

so far so good.

The amount of total possible outcomes = 2*2*2*2*2*2 = 64

And here is where i stumble.

The description is (6)
(4) = 15 Beneficial outcomes

And i am completely lost

what do they mean by (6)
(4) = 15

And how to calculate it? :confused:
Do you perhaps mean the following?

. . . . .\(\displaystyle \large{\binom{6}{4}}\)

Has your class not yet covered this notation? To learn about combinations and permutations, as well as the notations and computations related to them, try online lessons, such as here. ;)
 

Darketernal

New member
Joined
Mar 14, 2017
Messages
4


Sorry for the late reply, thank you so much for your help.

I didn't know how to express

(6)
(4)

in a proper manner, so it was very hard to find it online for me.
Could
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,464
I must be the worst person at math in the universe, oh well here goes.

My problem is that i don't understand the explanation (i don't know how to calculate it).

The probability question is as following:

You have 6 coins, what is the probability that you throw 4 times tails.

La place definition is the amount of beneficial outcomes divided by the amount of total possible outcomes.

so far so good.

The amount of total possible outcomes = 2*2*2*2*2*2 = 64

And here is where i stumble.

The description is (6)
(4) = 15 Beneficial outcomes

And i am completely lost

what do they mean by (6)
(4) = 15

And how to calculate it? :confused:
I am not sure that it is possible to explain even the rudiments of probability theory on a site like this. However, let's try to eliminate some darkness.

Some notation

FACTORIAL

4! = 4 * 3 * 2 * 1

3! = 3 * 2 * 1

2! = 2 * 1

1! = 1

A more formal definition is

\(\displaystyle 0! = 1\ and\)

\(\displaystyle n! = n * (n - 1)!\ for\ any\ n \in \mathbb N^+.\)

BINOMIAL COEFFICIENT

\(\displaystyle n,\ k \in \mathbb Z\ and\ 0 \le k \le n \implies \dbinom{n}{k} = \dfrac{n!}{k! * (n - k)!}.\)

Assume we flip a FAIR coin 6 times and write down the results as we get them.

As you saw there are 64 possibilities. And if the coin is fair, the probability of any one of them is

\(\displaystyle \dfrac{1}{64}.\)

I suggest that you write every possibility down in a systematic way, starting with HHHHHH and ending with TTTTTT. Count them up. You should have 64 and no duplicates.

How many cases will have 6 heads? Obviously 1. But

\(\displaystyle \dbinom{6}{6} = \dfrac{6!}{6! * (6 - 6)!} = \dfrac{1}{(6 - 6)!} = \dfrac{1}{0!} = \dfrac{1}{1} = 1.\)

How many cases will have 5 heads. It is not so obvious, but it is 6 cases because the T could come first, second, third, fourth, fifth, or sixth. And

\(\displaystyle \dbinom{6}{5} = \dfrac{6!}{5! * (6 - 5)!} = \dfrac{6 * 5 * 4 * 3 * 2 * 1}{5 * 4 * 3 * 2 * 1 * 1!} = \dfrac{6}{1} = 6.\)

How many cases will have 4 heads. I suggest you count them up. You will find if you have not made a mistake that you have 15. They will look like this:

HHHHTT 1
HHHTHT 2
HHTHHT 3
HTHHHT 4
THHHHT 5
HHHTTH 6
HHTHTH 7
HTHHTH 8
THHHTH 9
HHTTHH 10
HTHTHH 11
THHTHH 12
HTTHHH 13
THTHHH 14
TTHHHH 15

And

\(\displaystyle \dbinom{6}{4} = \dfrac{6!}{4! * (6 - 4)!} = \dfrac{6 * 5 * 4 * 3 * 2 * 1}{ 4 * 3 * 2 * 1 * 2!} = \dfrac{6 * 5}{2 * 1} = 15.\)
 
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Darketernal

New member
Joined
Mar 14, 2017
Messages
4


I don't know how to thank you. I especially want to thank you for the words 'clearing up the darkness' i definitely want you to keep up that strategy because it's exactly on this pivotal point where teachers are often not clear on how to correctly do it.

I mean basically it was 6!/4! = 30 then 6!-4! = 2! then 30/2! = 15

I really start understanding why i didn't understand it, because my teachers gave such a horrendous explanation, a far cry from your crystal clear explanation.

thank you so much again.
 
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