Solution: 2014 Winter Final - 14

Author: Michiel Smid

Question

We flip a fair coin repeatedly and independently, resulting in a sequence of heads ($H$) and tails ($T$). We stop flipping the coin as soon as this sequence contains $H$ or $TTTT$. What is the probability that this sequence contains at most two $T$s?

(a)

6/8

(b)

4/8

(c)

5/8

(d)

7/8

Solution

Let $A$ be the event that the sequence contains no $T$.
The set of outcomes for $A$ is $(H)$.
The probability of $A$ is $ Pr(A) = \frac{1}{2}$ since the coin has a 50% chance of landing heads.
Let $B$ be the event that the sequence contains exactly one $T$.
The set of outcomes for $B$ is $(HT)$.
The coin has a 50% chance of landing heads followed by 50% chance of landing tails.
The probability of $B$ is $ Pr(B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$
Let $C$ be the event that the sequence contains exactly two $T$s.
The set of outcomes for $C$ is $(HTT)$.
the coin has a 50% chance of landing heads followed by two 50% chances of landing tails.
The probability of $C$ is $ Pr(C) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$

The probability that the sequence contains at most two $T$s is the \sum of the probabilities of events $A$, $B$, and $C$:

$(Pr(\text{at most 2} T)) $

$ = Pr(A) + Pr(B) + Pr(C) $

$ = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}$