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?
$\endgroup$4 Answers
$\begingroup$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
$$ \implies P(A\cap B)=P(A)P(B). $$
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: $\frac{1}{2^5}$.
$\endgroup$ $\begingroup$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 $\frac1{32}$.
$\endgroup$ $\begingroup$I'll explain for two events, and you can extend the logic beyond to any number. Suppose you have an event with probability $m/n$. That is, in $n$ trials, you would expect the event to happen $m$ times. This is then equivalent to drawing a random number $1$ through $n$, and the event is getting a number $1$ through $m$.
So, when you draw a number, there are $n$ possibilities, and $m$ of them are the event. When you draw two numbers, there are $n$ possibilities for the first one, and another $n$ possibilities for the second one, so a total of $n^2$ possibilities of 2 numbers. Similarly, there are $m$ ways to get the event in the first trial, and another $m$ ways to get it in the second trial, so a total of $m^2$ ways to get the event. Thus the probability of the event is $m^2/n^2 = (m/n)^2$.
The events need to be independent for this to work, because you may not have $n$ possibilities or $m$ ways to get the event the second time. For example, if we don't put the number back when we draw again, the second time we draw there are $n-1$ possibilities, and either $m$ or $m-1$ ways to get the event, depending on if we got it in the first draw.
$\endgroup$ $\begingroup$The multiplication rule (at least one version of it) is stated as follows: if $A_1,\cdots, A_n$ are independent events, then $$\mathbf{P}(A_1\cap\cdots\cap A_n)=\mathbf{P}(A_1)\cdot\mathbf{P}(A_2)\cdot \cdots\cdot \mathbf{P}(A_n).$$ The addition rule is stated as follows: if $A_1,\cdots, A_n$ are disjoint events, then $$\mathbf{P}(A_1\cup A_2\cup\cdots\cup A_n)=\mathbf{P}(A_1)+\mathbf{P}(A_2)+\cdots+\mathbf{P}(A_n).$$ Thus to determine whether to use multiplication or addition rule, you want to determine if you are dealing with independent events happening altogether, in which case $A_1\cap\cdots \cap A_n$, or disjoint events at least one of which happens, in which case $A_1\cup\cdots\cup A_n$.
In your case, you want to compute the probability that all the 5 flips give a head, which belongs to the case of independent events happening altogether. Thus you should use the multiplication rule rather than the addition rule.
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