pipantasi4

Answered

2022-06-30

I know the independent concept and multiplication rules.

Example Flip Coin

5 times flip, how many times success of Head?

Ans : $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{32}$

Why do we use multiplication rules, not other rules?

I know if we use addition then probability 2.5 which is invalid because probability should be between 0 to 1. Same for Subtraction and division.

But, I am still confused why we use multiplication? What is the basic and core logic behind this?

Can anyone tell me the concept of this?

Example Flip Coin

5 times flip, how many times success of Head?

Ans : $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{32}$

Why do we use multiplication rules, not other rules?

I know if we use addition then probability 2.5 which is invalid because probability should be between 0 to 1. Same for Subtraction and division.

But, I am still confused why we use multiplication? What is the basic and core logic behind this?

Can anyone tell me the concept of this?

Answer & Explanation

trantegisis

Expert

2022-07-01Added 20 answers

Given two events $A$ and $B$, then the conditional probability of $A$ given $B$ (the probability of $A$ if we know that $B$ has occured) is defined as

$P(A|B)=\frac{P(A\cap B)}{P(B)},$

for this conditional probability, when the occurrence of $A$ is independent of $B$, then we have

$P(A|B)=P(A),$

it is clear from the two equations above two events $A$ and $B$ are defined to be statistically independent if

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P(A\cap B)=P(A)P(B).$

which means the information about the value of one event provides no information about the happening of the other event.

Now, suppose we consider the toss of 5 coins at the same time, the probability of occurrence of 5 heads $hhhhh$ is:

$P(hhhhh)=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}.$

here, you can consider that there are $2\times 2\times 2\times 2\times 2$ occurrences from which one is the case that we desired: $hhhhhh$, hence the probability for this case is: $\frac{1}{{2}^{5}}$.

$P(A|B)=\frac{P(A\cap B)}{P(B)},$

for this conditional probability, when the occurrence of $A$ is independent of $B$, then we have

$P(A|B)=P(A),$

it is clear from the two equations above two events $A$ and $B$ are defined to be statistically independent if

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P(A\cap B)=P(A)P(B).$

which means the information about the value of one event provides no information about the happening of the other event.

Now, suppose we consider the toss of 5 coins at the same time, the probability of occurrence of 5 heads $hhhhh$ is:

$P(hhhhh)=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}.$

here, you can consider that there are $2\times 2\times 2\times 2\times 2$ occurrences from which one is the case that we desired: $hhhhhh$, hence the probability for this case is: $\frac{1}{{2}^{5}}$.

Jamison Rios

Expert

2022-07-02Added 6 answers

Lets first simplify the problem to 1 flip and work from there. If we have one flip, we've got 2 options: h (heads) or t (tails).

If we throw 2 coins, we've got 4 options: ht,th,hh,tt all of which have an equal chance of happening.

Move to 3 coins and we get 8 options: hhh,hht,hth,thh,htt,tht,tth,ttt. Again, all of which have an equal chance of happening. It will be obvious that the different possibilities keep doubling because every throw has 2 options. If instead we roll a 6 sided die, we'd have to multiply the total amount of options by 6.

This being said, we still divide our "positive" result by all the options. So in your example the only positive result is hhhhh which means that there's 1 positive result in a total of 32 options. (You can look at it as an out branching tree.) Therefore the chance is $\frac{1}{32}$.

If we throw 2 coins, we've got 4 options: ht,th,hh,tt all of which have an equal chance of happening.

Move to 3 coins and we get 8 options: hhh,hht,hth,thh,htt,tht,tth,ttt. Again, all of which have an equal chance of happening. It will be obvious that the different possibilities keep doubling because every throw has 2 options. If instead we roll a 6 sided die, we'd have to multiply the total amount of options by 6.

This being said, we still divide our "positive" result by all the options. So in your example the only positive result is hhhhh which means that there's 1 positive result in a total of 32 options. (You can look at it as an out branching tree.) Therefore the chance is $\frac{1}{32}$.

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