# Math Final Tomorrow! :S Probability Question.

• Jun 16th 2009, 06:41 PM
facesonfilm
Math Final Tomorrow! :S Probability Question.
okay so I have my math final tomorrow and he said that this question was on the test, I've tried all the approaches I could come up with but still haven't gotten the answer.

The question is:

Two hockey teams, the Spartans and the Burners, hav a sudden death penalty shootout to decide who wins the game. The teams take penalty shots in turns. The first team to score wins. The probability that the Spartans score on any penalty shot is 0.3 and the probability that the Burners score on any penalty shot is 0.4. If the Spartans take the first penalty shot, determine the probability that they win the game.

thanks so much !
• Jun 16th 2009, 07:27 PM
Soroban
Hello, facesonfilm!

This involves an infinite series . . .

Quote:

Two hockey teams, the Spartans and the Burners, have a sudden-death
penalty shootout to decide who wins the game.
The teams take penalty shots in turns. The first team to score wins.
The probability that the Spartans score on any penalty shot is 0.3
and the probability that the Burners score on any penalty shot is 0.4.

If the Spartans take the first penalty shot,
determine the probability that they win the game.

Consider the following scenarios . . .

Spartans win on their 1st try:
. . $P(\text{S wins, 1st try}) \:=\:0.3$

Spartans win on their 2nd try:
. . They miss their 1st try: . $0.7$
. . The Burners miss their 1st try: . $0.6$
. . The Spartans make their 2nd try: . $0.3$
Spartans win on their 2nd try:
. . $P(\text{S wins, 2nd try}) \:=\:(0.7)(0.6)(0.3) \:=\:(0.3)(0.42)$

Spartans win on their 3rd try:
. . They miss their 1st try: . $0.7$
. . The Burners miss their 1st try: . $0.6$
. . The Spartans miss their 2nd try: . $0.7$
. . The Burners miss their 2nd try: . $0.6$
. . The Spartans make their 3rd try: . $0.3$
Sparts win on their 3rd try:
. . $P(\text{S wins, 3rd try}) \:=\:(0.7)(0.6)(0.7)(0.6)(0.3) \:=\:(0.3)(0.42)^2$

Skipping the listing, we get:
. . $P(\text{S wins, 4th try}) \:=\:(0.3)(0.42)^3$

The probability that the Spartans win on their 1st try or their 2nd try
. . or their 3rd try or their 4th try, etc. is:

$P(\text{Spartans win}) \;=\;0.3 + (0.3)(0.42) + (0.3)(0.42)^2 + (0.3)(0.42)^3 + \hdots$

. . . . . . . . . . . $= \;0.3\underbrace{\bigg[1 + 0.42 + (0.42)^2 + (0.42)^3 + \hdots\bigg]}_{\text{geometric series}}$

The sum of the geometric series is: . $\frac{1}{1-0.42} \:=\:\frac{1}{0.58}$

Therefore: . $P(\text{Spartans win}) \;=\;(0.3)\left(\frac{1}{0.58}\right) \:=\:\frac{0.30}{0.58} \:=\:\frac{15}{29}$