Probability 3

1. Feb 18, 2006

Natasha1

Could anyone please help me with making a start.... and a finish to this little exercise. Many thanks Nat.

Four people A, B, C and D are to play in a small table-tennis tournament played on a simple knock-out basis: their names are drawn at random to play in two pairs, then the two winners play in the final. The probability that a player beats a player with a later letter is 2/3. All matches are independent. Find the probabilities that:

1) A wins the tournament

2) C and D meet in the final

3) B and C meet at some stage

Must I use a tree diagram?

2. Feb 18, 2006

arildno

1) A must win two fights, irrespective of the pairing arrangements. What is the probability of that?

2) This does depend on the probability that C and D does NOT meet each other in the first fight, and that BOTH win their first fight (irrespective of who beats A or B)

3. Feb 18, 2006

Natasha1

Is 1) p(A) = 2/3 * 2/3 = 4/9

4. Feb 18, 2006

arildno

Correct!

5. Feb 18, 2006

Natasha1

Is 2) p(C & D to meet in final) = 2/3 * 2/3 * 2/3 * 2/3 = 16/81

6. Feb 18, 2006

arildno

No.
1. There is 2/3 chance they'll not meet in the first round. 1/3 of those times, C will go to the finals, and 1/3 of those times again D will also go to the finals to meet C.

7. Feb 18, 2006

Natasha1

I see so...

P(C&D meet in final) = 2/3 * 1/3 * 1/3 = 2/27

8. Feb 18, 2006

arildno

3*3*3 equals 27 last time I checked..

9. Feb 18, 2006

Natasha1

lol, i did spot it straight away and corrected it :shy:

Right then...

for number 3

Last edited: Feb 18, 2006
10. Feb 18, 2006

arildno

Now, for 3) calculate separately the probabilities for the disjoint events a) B&C meets in the first round b) B&C meet in the finals.

Either a) or b) may happen..

11. Feb 18, 2006

Natasha1

a) p(B&C meets in the first round) = 1/3

b) p(B&C meet in the finals) = 2/27

So the p(B&C meet at some stage) = 1/3 * 2/27 = 2/81

12. Feb 18, 2006

arildno

Incorrect!
a) is correct, but why do you think b) is correct??

Furthermore, you are to have probabilities of disjoint events EITHER of which occuring will mean that b&c meets at some stage. Should you multiply the probabilities together in that case?

Last edited: Feb 18, 2006
13. Feb 18, 2006

Natasha1

Ok so

a) p(B&C meets in the first round) = 1/3

b) p(B&C meet in the finals) = ?
Well I thought this was correct as for question b) the p(C&D to meet in final is 2/27 so why would p(B&C meet in final) would be different?

So the p(B&C meet at some stage) = 1/3 + ? = ?

14. Feb 18, 2006

arildno

b) will be different in that in the first case, for example, C must always beat a BETTER player in order to go to the finals.
There is 1/3 chance that the initial pairing is (A&B, C&D) What is the chance that this was the set up AND that both B and C wins (and thus proceed to the finals to meet each other)?

Furthermore, it is a 1/3 chance that the initial pairing was (A&C,B&D) What is the chance that this was the set up AND that both B and C wins
(and thus proceed to the finals to meet each other)?

15. Feb 18, 2006

Natasha1

b) P = 2/3 * 1/3 * 1/3 * 1/3 = 2/81

So p(B&C meet at some stage) = 1/3 + 2/81 = 29/81

16. Feb 18, 2006

arildno

No, no, no!

17. Feb 18, 2006

Natasha1

Let me see...

There is 1/3 chance that the initial pairing is (A&B, C&D) What is the chance that this was the set up AND that both B and C wins (and thus proceed to the finals to meet each other)? this gives 1/3 * 1/3 * 1/3 = 1/27

Furthermore, it is a 1/3 chance that the initial pairing was (A&C,B&D) What is the chance that this was the set up AND that both B and C wins
(and thus proceed to the finals to meet each other)? this gives 1/3 * 1/3 * 2/3 = 2/27

18. Feb 18, 2006

arildno

Almost correct, but the probability for the first case is also 2/27, since C beats D 2/3 of the time.

Thus, the TOTAL probability of B&C meeting in the finals is 2/27+2/27=4/27

And the total probability of B&C meeting at some stage is therefore....?

19. Feb 18, 2006

Natasha1

p(B&C meeting at some stage) = 1/3 + 4/27 = 13/27

20. Feb 18, 2006

You're done!