pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,976
While I have carefully read this ongoing thread, I have stayed out of it because the instructor’s clearly thinks the denominators should be 6! So to correct that mistake may mean that you lose valuable points on this project. You will have to decide for yourself which way to go.
But “deep in Gene’s heart” is the correct answer!
For a little history on this see: http://www.freemathhelp.com/forum/viewtopic.php?t=9767.
It does indeed depend upon the total possible games even if they not played.
Melissa, you yourself wrote “I multiply the branches on a tree diagram it gives me different results.” Because the teams are evenly matched, no one can deny that the probability that Comets will win the first two is (1/4). How does the six square with that? Now consider the question: “The probability that one of the team wins 2 straight games is 2/6” Does that mean ‘win the first two’ or ‘any two straight’? The probability that the first two matches will be won be the same team is (1/4)+(1/4)=(1/2), right? [P(CC or BB)=P(CC)+P(BB) because they are disjoint.] Again where is the six? Look at Eliz’s first response: “Possible outcomes: BBB, BBC, BCB, BCC, CBB, CBC, CCB, and CCC. Of these, only BBB, BBC, CCB, and CCC include the first two are won by the same team.” That agrees with the (1/2).
Again, instructor’s clearly thinks the denominators should be 6!
But Eliz’s first response and Gene’s ‘deep-heart’ are correct!
You will have to decide for yourself which way to go.
If it makes any difference, I have taught this for twenty-five years.
But “deep in Gene’s heart” is the correct answer!
For a little history on this see: http://www.freemathhelp.com/forum/viewtopic.php?t=9767.
It does indeed depend upon the total possible games even if they not played.
Melissa, you yourself wrote “I multiply the branches on a tree diagram it gives me different results.” Because the teams are evenly matched, no one can deny that the probability that Comets will win the first two is (1/4). How does the six square with that? Now consider the question: “The probability that one of the team wins 2 straight games is 2/6” Does that mean ‘win the first two’ or ‘any two straight’? The probability that the first two matches will be won be the same team is (1/4)+(1/4)=(1/2), right? [P(CC or BB)=P(CC)+P(BB) because they are disjoint.] Again where is the six? Look at Eliz’s first response: “Possible outcomes: BBB, BBC, BCB, BCC, CBB, CBC, CCB, and CCC. Of these, only BBB, BBC, CCB, and CCC include the first two are won by the same team.” That agrees with the (1/2).
Again, instructor’s clearly thinks the denominators should be 6!
But Eliz’s first response and Gene’s ‘deep-heart’ are correct!
You will have to decide for yourself which way to go.
If it makes any difference, I have taught this for twenty-five years.
. I will post Tuesday night how I did for anyone who is curious what my instructor thinks lol.