Answer:
[tex]P(W') = \frac{6}{n+6}[/tex]
Step-by-step explanation:
Let P(W) represents the probability that Paul wins
Let P(W') represents the probability that Paul does not win
Given
[tex]P(W) = \frac{n}{n+6}[/tex]
Required
[tex]P(W')[/tex]
In probability, the sum of opposite probability equals 1;
This implies that
[tex]P(W) + P(W') = 1[/tex]
Substitute [tex]P(W) = \frac{n}{n+6}[/tex] in the above equation
[tex]P(W) + P(W') = 1[/tex] becomes
[tex]\frac{n}{n+6}+ P(W') = 1[/tex]
Subtract [tex]\frac{n}{n+6}[/tex] from both sides
[tex]\frac{n}{n+6} - \frac{n}{n+6} + P(W') = 1 - \frac{n}{n+6}[/tex]
[tex]P(W') = 1 - \frac{n}{n+6}[/tex]
Solve fraction (start by taking the LCM)
[tex]P(W') = \frac{n + 6 - n}{n+6}[/tex]
[tex]P(W') = \frac{n - n + 6}{n+6}[/tex]
[tex]P(W') = \frac{6}{n+6}[/tex]
Hence, the probability that Paul doesn't win is [tex]P(W') = \frac{6}{n+6}[/tex]
