# Why Binomial Distribution formula includes the “not-happening” probability?

Suppose I have a dice with 6 sides, and I let a random variable $X$ be the number of times I get 3 points when I throw the dice.

So I throw the dice for $10$ times, I want to find the probability of getting 3 points from the dice for $4$ times, ie: $P(X=4)$.

Since the order doesn't matter, there are $\binom{10}{4}=210$ ways and the chance of getting a 3 point is $\frac { 1 }{ 6 }$. Also, because I want to have $4$ of such occurrence, it would be $\frac{1}{6}^4$. So, I could just calculate $P(X=4)=\binom{10}{4}\frac { 1 }{ 6 }^4 =0.027006173\approx 2.7\%$.

But, suppose if I use the Binomial Distribution formula, it would be a little different because it needs to multiply the "not-happening" probability to it. The Binomial Distribution looks like this: $$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

So if I plug in my values, it would be: $$P(X=4)=\binom{10}{4}(\frac{1}{6})^4(\frac{5}{6})^{6}=0.054=5.4\%$$

Here, $2.1\%$ is lesser than $5.4\%$. What's the difference between the two values? Which is the correct value?

Intuitively, I find the Binomial Distribution may be more accurate since it dictates the situation to consider both the happening and not-happening outcomes. But usually, I thought we just multiply the probabilities of what we want it to happen as long as the events are independent. So the first method sounds quite okay too. Eg: What's the probability to get 2 heads out of 5 flips of a fair coin, I just use $\frac { 1}{ 2} \times \frac { 1}{ 2}$. The not-happening probabilities are not cared of.

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To get exactly 4 "3 points", the other 6 throws have to not be "3 points". The other 6 throws not being 3 points is also part of what you want to happen if exactly 4 of the 10 throws are 3 points. So, you need to multiply by $(5/6)^6$. By the way, in your second displayed equation, you should have ${10\choose 4}(1/6)^4(5/6)^6$ – David Mitra Feb 18 '12 at 16:33
oh yea, thanks. It should ${10\choose 4}(1/6)^4(5/6)^6$ in the second line. Updated the equation. If I need to multiply the "not-happening" probability, then suppose if I want to find the probability of getting 2 heads out of 5 flip of a fair coin, it wouldn't it be $\binom{5}{2}{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }=\binom{5}{2}{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }$? But shouldn't it be just $\frac { 1}{ 2} \times \frac { 1}{ 2}$? – xenon Feb 18 '12 at 16:51
No, not just $(1/2)(1/2)$ for exactly two heads. You $need$ the other flips to not be heads; which means you multiply by $(1/2)^3$. – David Mitra Feb 18 '12 at 17:10
@xEnOn: The calculation that gave about $5.4$% is based on the correct analysis, and the calculator work is right. The first calculation, the one that gave about $2.7$%, is based on an incorrect analysis (formula). In addition, there was a calculator mistake. The wrong formula gives an answer of about $16.2$%, not $2.7$%. This is clear if you compare the two expressions. The second multiplies the first by $(5/6)^6$, which is less than $1$. – André Nicolas Feb 18 '12 at 17:18
@xEnOn : Please don't write $\frac 16 ^4$ if you mean $\left(\frac16\right)^4$. The former expression should be used only if you mean $(1^6)/4$. – Michael Hardy Feb 18 '12 at 17:53

Your first argument is a bit off. $X=4$ means exactly four of the tosses resulted in $3$-points To get exactly four $3$-points, the other six throws have to not be $3$- points. The other six throws not being $3$-points is also part of what you want to happen if exactly four of the 10 throws are $3$-points.

Look at a particular case: the first four throws are 3-points and there are exactly four 3-points. Then the throw sequence was $$3\text{-pt}\ \ 3\text{-pt}\ \ 3\text{-pt}\ \ 3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \$$ The probability of this occurring is $(1/6)^4(5/6)^6$.

So, you need to multiply your proposed answer by $(5/6)^6$. So, the correct answer is what is given by the Binomial distribution: ${10\choose4}(1/6)4(5/6)6$.

In your second argument (of the original post), you are correct assuming that you only drew two cards. If you drew three cards, the probability that exactly two were hearts would be $$\underbrace{{13\over52}{12\over 51}{39\over 50}}_{ \text{two hearts, then non heart}}+ \underbrace{{13\over52}{39\over 51}{12\over 50}}_{ \text{ heart, non heart, heart}}+ \underbrace{{39\over52}{13\over 51}{12\over 50}}_{ \text{ non heart, then two hearts}}$$

To be brief you do care about the "not happening" probabilities when you consider the probability that an event happens exactly $n$ times. Knowing that something does not happen is concrete information. In the die example, "not 3-pt" means that six of the flips were 1,2,4,5, or 6 points.

In your coin example, if you want the probability of exactly two heads in five flips, then the three non-head flips must be tails. What is the probability of flipping $H\ H\ T\ T\ T$? It's not $1/4$... It is $1/32$, as you should be able to see from the multiplication rule for a sequence of independent events.

The number of ways in which you can have exactly two heads in five flips is ${5\choose2}=10$ (it's the number of ways to choose two slots from five in which to put the two heads in; note the other three slots must be tails). The probability of exactly two heads in five flips is ${5\choose2}\cdot(1/2)^2(1/2)^3={10\over 32}={5\over16}$.

To see more concretely why your argument fails, consider this simple case. Toss a fair coin three times. What is the probability of having exactly one head? By your reasoning it is $3\cdot(1/2)^1>1$; which is nonsense. Perhaps it's just $1/2$? This is incorrect also:

The equally likely outcomes here are: $$HHH \ \ \ HHT \ \ \ HTT\ \ \ HTH$$ $$THH \ \ \ THT \ \ \ TTT\ \ \ TTH$$ And we see the probability of exactly one head is ${3\over8}$, which is exactly what the Binomial formuls gives: ${3\over8}={3\choose2}(1/2)^1(1/2)^2$.

Incidentally, you're proposed method would not even give the probability of having at least one head. Here, that probability is $7/8$.

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