# Question bb618

Apr 25, 2017

a) ~~42.86%
b) ~~41.98%
c) ~~39.36%

#### Explanation:

Probability of winning the first game means he must win the first and the outcome of winning or losing beyond that is not meaningful so let's think about possible outcomes

• $w 1 =$ win or lose
• $w 2 =$ win or lose
• $w 3 =$ win or lose

so in the case that w1=win then w2 and w3 can be a win or lose. This means that the total outcomes is

$p \left(w 1 = \text{win" and w2="win" and w3="win}\right) = \frac{3}{7} \cdot \frac{3}{7} \cdot \frac{3}{7}$

let's just say

$P \left(w 1 = w , w 2 = w , w 3 = w\right) = {\left(\setminus \frac{3}{7}\right)}^{3}$

another outcome is
$P \left(w 1 = w , w 2 = w , w 3 = l\right) = \setminus \frac{3}{7} \cdot \setminus \frac{3}{7} \cdot \setminus \frac{4}{7}$

continuing this line of thinking the other two outcomes where w1 wins is thus

$P \left(w 1 = w , w 2 = l , w 3 = l\right) = \setminus \frac{3}{7} \cdot \setminus \frac{4}{7} \cdot \setminus \frac{4}{7}$
and
$P \left(w 1 = w , w 2 = l , w 3 = w\right) = \setminus \frac{3}{7} \cdot \setminus \frac{4}{7} \cdot \setminus \frac{3}{7}$

now if we want to determine winning the first game we can consider the 4 outcomes or $P \left(S 1 \setminus \mathmr{and} S 2 \setminus \mathmr{and} S 3 \mathmr{and} S 4\right)$, where $S$ stands for scenario. You will notice there are 4 scenarios so. since these are all mutually exclusive we should be able to add the probabilities thus

a) =\frac{27}{343} + \frac{36}{343} + \frac{36}{343} +\frac{48}{343} ~~ 42.86%
you notice this is the same result for probability of winning a game. This makes sense because if we were to construct a tree you would see that winning the first game is always $\frac{3}{7}$ regardless of the path after it, but its a good exercise to see this.

the probability of winning the game exactly once is the same as
$P \left(w 1 = w , w 2 = l , w 3 = l\right) \mathmr{and} P \left(w 1 = l , w 2 = w , w 3 = l\right) \mathmr{and} P \left(w 1 = l , w 2 = l , w 3 = w\right)$.

b = 3* \frac{48}{343} ~~41.98% 

the probability of winning 2 out of 3 games the same as
$P \left(w 1 = w , w 2 = w , w 3 = l\right) \mathmr{and} P \left(w 1 = l , w 2 = w , w 3 = w\right) \mathmr{and} P \left(w = w , w 2 = l , w 3 = w\right) \mathmr{and} P \left(w 1 = w , w 2 = w , w 3 = w\right)$.

c = 3* \frac{36}{343} +\frac{27}{343} ~~39.36% #

it might not seem intuitive that we include the scenario where we win in all 3 games but this is because this is a valid outcome for winning 2 out of 3 times and the question is not exactly 2 out of 3 times