How to translate the conditional probability formulas into numerical values?

Question

I have some confusions regarding how to translate the conditional probability formulas here.

The problem goes like this: Suppose we have 10 independent tosses of a biased coin with $P(Heads) = p$. What is probability of 5 heads in the first 8 tosses and 3 heads in the last 5 tosses?

The given solution is as below:

Let G be the event “5 heads in the first 8 tosses” and let H be the event “3 heads in the last 5 tosses”.

I can understand the last part of each condition, but I am not sure how to derive those values in the red boxes. Any explanation is appreciated! thanks!

Answer 1

There are 10 tosses.

$G$ is a condition on the eight first tosses and $H$ on 5 five last ones. Therefore, the two could intersect and it's natural to understand each case separately.

First case scenario : only one head in the tosses $6-8$.

In that case, an event is in $G\cap H$ iff:

• There are 4 heads in tosses 1-5 : this occurs with probability $\binom{5}{4}p^4(1-p)$;

• Two heads in 9-10 : this occurs with probability p^2.

The coin might be biased but these two events are independant: therefore $$P(G\cap H|\mathrm{1~ head~ in~ tosses~ 6-8})=\binom{5}{4}p^4(1-p)\times p^2.$$

The two other cases are similar.

Edit : for the third case, (three heads in 6-8), an event is in $G\cap H$ iff:

• There are $5-3=2$ heads in tosses 1-5 : this occurs with probability $\binom{5}{2}p^2(1-p)^3$;

• $3-3=0$ heads in 9-10 : this occurs with probability $\binom{2}{0}p^0(1-p)^2=(1-p)^2$.

Answer 2

So I'm a bit unsure what you're asking. The expression in the red boxes come from the binomial distribution. If the probability for success $p$ is fixed and you are conducting $n$ independent trials, the binomial distribution calculates the probability that $r$ of these are successes. In this case you have three scenarios for satisfying the specified criterion:

• 4 heads in tosses 1-5 and 1 head in tosses 6-8 and 2 heads in tosses 9-10
• 3 heads in tosses 1-5 and 2 heads in tosses 6-8 and 1 heads in tosses 9-10
• 2 heads in tosses 1-5 and 3 heads in tosses 6-8 and 0 heads in tosses 9-10

as the toss groups 1-5, 6-8 and 9-10 are independent the probability for these can be calculated as

• $b(4, 5, p)b(1, 3, p)b(2, 2, p)$
• $b(3, 5, p)b(2, 3, p)b(1, 2, p)$
• $b(5, 2, p)b(3, 3, p)b(0, 2, p)$

where $b(r, n, p)$ is the binomial distribution. You will find that this yields the expression shown.

Since you have outlined only part of the red boxed I thought you might be also asking how to calculate the nCr E.g $ {5 \choose 4}$, ${3 \choose 1}$ and so on? These can be computed as $$ {5 \choose 4} = \frac{5!}{(5 - 4)!4!} $$ or in general $$ {n \choose r} = \frac{n!}{(n-r)!r!} $$ The nCr (n choose r) is an important size in statistics as it describes how many ways you can combine $r$ elements from a total of $n$ unique elements, meaning you don't care about the ordering of the $r$ elements. As this is so important, most calculators have a button (often marked nCr) for calculating the equation shown above.

Answer 3

Alternative route:

If $X$ denotes the number of heads in tosses $1,2,3,4,5$ and $Y$ denotes the number of heads in tosses $6,7,8$ and $Z$ the number of heads in tosses $9,10$ then to be found is: $$\sum_{k}P(X=5-k,Y=k,Z=3-k)$$

Here $X,Y,Z$ are independent with $X\sim\mathsf{Binom}(5,p)$, $Y\sim\mathsf{Binom}(3,p)$ and $Z\sim\mathsf{Binom}(2,p)$.

This leads to:$$\cdots=\sum_{k=1}^3P(X=5-k)P(Y=k)P(Z=3-k)=$$$$\sum_{k=1}^3\binom5{5-k}p^{5-k}(1-p)^k\binom3kp^k(1-p)^{3-k}\binom2{3-k}p^{3-k}(1-p)^{k-1}=$$$$\sum_{k=1}^3\binom5{5-k}\binom3k\binom2{3-k}p^{8-k}(1-p)^{2+k}$$