Question

Question #bb618

Answer 1

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

• `w1 =` win or lose
• `w2 =` win or lose
• `w3 =` 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(w1="win" and w2="win" and w3="win") = 3/7 * 3/7* 3/7`

let's just say

`P(w1=w,w2=w,w3=w) =(\frac{3}{7})^3`

another outcome is
`P(w1=w,w2=w,w3=l) =\frac{3}{7} * \frac{3}{7}*\frac{4}{7}`

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

`P(w1=w,w2=l,w3=l) =\frac{3}{7} * \frac{4}{7} * \frac{4}{7}`
and
`P(w1=w,w2=l,w3=w) =\frac{3}{7} * \frac{4}{7} * \frac{3}{7}`

now if we want to determine winning the first game we can consider the 4 outcomes or `P(S1 \or S2 \or S3 or S4 )`, 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 `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(w1=w,w2=l,w3=l) or P(w1=l,w2=w,w3=l) or P(w1=l,w2=l,w3=w)`.

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

the probability of winning 2 out of 3 games the same as
`P(w1=w,w2=w,w3=l) or P(w1=l,w2=w,w3=w) or P(w=w,w2=l,w3=w) or P(w1=w,w2=w,w3=w)`.

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