Question 5 of 12

Question 5

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two teams, each independently winning any given game with probability $p=\tfrac12$ (evenly matched); series stops the instant either team reaches 4 wins.

Find. P(series decided at game 4), P(at game 5), P(at game 6), P(reaches/goes to game 7).

Approach. "Ends exactly at game k" requires one team to have exactly 3 wins in the first $k-1$ games and then win game k; count both teams and both orderings with the binomial coefficient, then divide by $2^k$.

1. (a) Ends after 4 games. One team must win all 4 games. Either team can be the sweeper: $$P(4)=2\times\left(\tfrac12\right)^4=\boxed{\dfrac18=0.125}$$
2. (b) Ends after 5 games. The eventual winner must take exactly 3 of the first 4 games ($\binom43$ orderings) and then win game 5; either team can be the winner: $$P(5)=2\times\binom43\left(\tfrac12\right)^5=2\times4\times\tfrac1{32}=\boxed{\dfrac14=0.25}$$
3. (c) Ends after 6 games. The winner takes exactly 3 of the first 5 games ($\binom53$) then wins game 6: $$P(6)=2\times\binom53\left(\tfrac12\right)^6=2\times10\times\tfrac1{64}=\boxed{\dfrac{5}{16}=0.3125}$$
4. (d) Goes to a 7th game. Equivalent to "series is 3–3 after 6 games" (both teams have won exactly 3): $\binom63$ orderings out of $2^6$: $$P(7)=\binom63\left(\tfrac12\right)^6=\dfrac{20}{64}=\boxed{\dfrac{5}{16}=0.3125}$$ Cross-check: $P(4)+P(5)+P(6)+P(7)=\tfrac18+\tfrac14+\tfrac5{16}+\tfrac5{16}=\tfrac{2+4+5+5}{16}=1$, confirming the four outcomes exhaust the sample space.