Tennis Match of Probability Part 1

There are 4 ways Peter can win the match:

1. Jacob wins 0 games

2. Jacob wins 1 game

3. Jacob wins 2 games

4. Jacob wins 3 games

The probability of Peter winning 4 games with Jacob winning 0 games is ${(0.7)}^4$ .

If Peter wins the match while Jacob wins 1 game, then 5 games are played in total. The last game has to be won by Peter, so there are 4 possibilities for the game Jacob won. Therefore, the probability of Peter winning the match with Jacob winning 1 game is $4 \times 0.3 \times {(0.7)}^4$ .

If Peter wins the match while Jacob wins 2 games, then 6 games are played in total. The last game has to be won by Peter, so there are ${5 \choose 2} =10$ possibilities for the 2 games Jacob won. Therefore, the probability of Peter winning the match with Jacob winning 2 games is $10 \times {(0.3)}^2 \times {(0.7)}^4$ .

If Peter wins the match while Jacob wins 3 games, then 7 games are played in total. The last game has to be won by Peter, so there are ${6 \choose 3} =20$ possibilities for the 3 games won by Jacob. Therefore, the probability of Peter winning the match with Jacob winning 3 games is $20 \times {(0.3)}^3 \times {(0.7)}^4$ .

To find the probability of Peter winning the match, add the 4 numbers obtained together. For simplicity, you can factor out the ${(0.7)}^4$ , to make the sum of the 4 numbers ${(0.7)}^4 \times [1+4 \times 0.3+10 \times {(0.3)}^2+20 \times {(0.3)}^3]$ .

The final result is 0.873964, which is $\boxed {87.3964}$ as a percentage.