Answer
$\frac{15}{4}$
Work Step by Step
If we flip a fair coin, then we have $1$ chance in $2$ of tossing heads $H$ and $1$ chance in $2$ of tossing tails $T$.
$$P(H)=\frac{1}{2}=0.5$$
$$P(T)=\frac{1}{2}=0.5$$
We flip the coin until we obtained two tails or until we have flipped the coin $3$ times.
let $X$ be the number of flips of the coin.
Case 1- $X=2$
The first two flips are then both tails:
$$P(X=2)=P(T)P(T)=(0.5)^2=0.25$$
Case 2- $X=3$
$1$ of the first $2$ flips has to contain heads and the other two flips have to be tails.
$$P(X=3)=^2C_1 (P(T))^2P(H)=2(0.5)^2(0.5)=2(0.5)^3=0.25$$
Case 3- $X=4$
$2$ of the first $3$ flips has to contain heads and the other two flips have to be tails.
$$P(X=4)=^3C_2 (P(T))^2(P(H))^2=3(0.5)^2(0.5)^2=3(0.5)^4=0.1875$$
Case 4- $X=5$
$3$ of the first $4$ flips has to contain heads and the other two flips have to be tails.
$$P(X=5)=^4C_3 (P(T))^2(P(H))^3=4(0.5)^2(0.5)^3=4(0.5)^5=0.125$$
Case 5- $X=6$
$4$ or $5$ of the first has to contain heads (as there can be at most one tail among the first $5$ flips when $X=6$). The $6_{th}$ flip could be either heads or tails, as we stop flipping the coins after $6$ flips ( no matter the number of tails).
$P(X=6)=^5C_4 (P(T)(P(H))^4+ ^5C_5(P(H))^5$
$=5(0.5)(0.5)^4+5(0.5)^5$
$=5(0.5)^5+(0.5)^5$
$=6(0.5)^5$
$=0.1875$
$\underline{Expected Value}$
The expected value (or mean) is the sum of the product of each probability of each possibility $x$ with its probability $P(x)$.
$E(x)=sum xP(x)$
$= 2(0.25)+3(0.25)+4(0.1875)+5(0.125)+6(0.1875 )$
$=3.75$
$=\frac{15}{4}$
