Solution: 2014 Fall 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 one $H$ or eight $T$s. What is the probability that this sequence contains at most 7 $T$s?

(a)

$\sum_{k=0}^{7} (1/2)^{k+1}$

(b)

$1 - (1/2)^{7}$

(c)

None of the above.

(d)

$\sum_{k=0}^{7} (1/2)^{k}$

Solution

Let’s go through each option and explain

$1 - (1/2)^{7}$
This is just the complement rule, which usually works but they're subtracting the probability that we get 6 tails followed by a head
It might've worked if they were subtracting the probability that we get 7 tails followed by a head from the whole set of possibilities
$\sum_{k=0}^{7} (1/2)^{k}$
This one is wrong because it starts at $ { left( \frac{1}{2} \right)}^0 $ but it doesn't take into account the fact that we have to flip a head to stop, which has a probability of $ \frac{1}{2} $
$\sum_{k=0}^{7} (1/2)^{k+1}$
This one is beautiful
It says, probability of flipping a heads right away, probability of flipping a tails folloewd by a heads, probability of flipping 2 tails followed by a heads, and so on until 7 tails followed by a heads