Translate the conditional probability formulas into numerical values

conditional probabilityprobability

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”.

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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!

Best Answer

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 $\binom{2}{0}p^0(1-p)^2=(1-p)^2$.

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