Extension Qs- Exercise 10H of the Cambridge Year 12 3 unit textbook (1 Viewer)

[bleh1234]

I have no idea how to do the two of these questions D: Any help is appreciated!
26) During the seven games of the football season, Max and Bert each miss three consecutive games. The games to be missed by each player are randomly and independently selected.
a) What is the probability that they both have the first game off together? 1/25
b) What is the probability that the second game is the first one missed by both players? 3/25
c) What is the probability that Max and Bert miss at least one of the same games? 19/25
27) Eight players make the quarter-finals at Wimbledon. The winner of each of the quarter finals plays a semi-final to see who enters enters the final.
a) Assuming that all eight players are equally likely to win a match, show that the probability that any two particular players will play each other is 1/4.
b) What is the probability that two particular people will play each other if the tournament starts with 16 players? 1/8
c) What is the probability that two particular players will meet in a similar knockout tournament if 2^n players enter? 2^(1-n)
I HATE PERMS AND COMBSSSSSSSSSSS!!!!!!!!!!!!!!!!!!!!!!!!1

 

[Ambility]

For 26a:

Think about which games a player could miss. They have to be 3 games in a row of 7 games. That means they could miss either of the following combinations:

(1,2,3)
(2,3,4)
(3,4,5)
(4,5,6)
(5,6,7)

Only one of those combinations is where the player has the first game off. That's a probability that a player will have the first game off. For two players, you square this giving .

 

[bleh1234]

Thank you very much

 

[braintic]

I HATE PERMS AND COMBS

There is no perms and combs required here.

For the first question, draw up a grid as you would for rolling 2 dice. But instead of labelling the rows and columns 1 to 6, label them with the outcomes given in Ambility's post. That is, a 5 by 5 grid. The outcomes across the top are Max's; the outcomes down the side are Bert's. Then tick the boxes that answer the question.

27a:

In a one round (2 player) competition, there is 100% chance they meet.

In a two round (4 player) competition, they can either
(I) meet in the 1st round (1/3, since each player has 3 possible opponents); OR
(II) NOT meet in first round (2/3) AND both win 1st round match (1/2 times 1/2) AND then meet in ensuing 2 player competition (100%)
1/3 + 2/3 × 1/2 × 1/2 × 1 = 1/2

In a three round (8 player) competition, they can either
(I) meet in the 1st round (1/7, since each player has 7 possible opponents); OR
(II) NOT meet in first round (6/7) AND both win 1st round match (1/2 times 1/2) AND then meet in ensuing 4 player competition (1/2)
1/7 + 6/7 × 1/2 × 1/2 × 1/2 = 1/4

See if you can extend that logic to 16 players