Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game
, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?
1 answer:
Answer:
Probability is
Step-by-step explanation:
This problem can be easily done tree diagram.
Few things keep in mind :
- From starting of match three victories of A.
- Less then or equal to two victories of B.
- In each matches there are two possibilities either favourable for A or B.
First case if A wins first match
- two victories of B then two of A
- victory of B then A then B then A
- victory of B then A then A
Second case if B wins first match(favorable)
- victory of two B then three A
So total cases are 4 and favorable is one
probability=
[Tree diagram is in attachment]
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Answer:
See below for graphs
- (2, -3), r = 2 (blue)
- (-3, -1), r = 1 (green)
- (3, -1), r = 3 (purple)
- (4, 1), r = 3 (red)
Step-by-step explanation:
I like to solve a set of problems like this once, then use a calculator or spreadsheet to crunch the numbers.
We can start with the general form of the equation for a circle:
x^2 +y^2 +ax +by +c = 0
Completing the square gives ...
(x^2 +ax +(a/2)^2) +(y^2 +by +(b/2)^2) = (a/2)^2 +(b/2)^2 -c
(x +(a/2))^2 +(y +(b/2))^2 = (a/2)^2 +(b/2)^2 -c
Comparing this to the standard form formula ...
(x -h)^2 + (y -k)^2 = r^2
we see the relations are ...
- h = -a/2
- k = -b/2
- r = √((a/2)² +(b/2)² -c)
where (h, k) is the center and r is the radius.
Then your circle centers and radii are ...
- (h, k) = (-(-4)/2, -6/2) = (2, -3); r = √(4+9-9) = 2
- (h, k) = (-6/2, -2/2) = (-3, -1); r = √(9+1-9) = 1
- (h, k) = (-(-6)/2, -2/2) = (3, -1); r = √(9+1-1) = 3
- (h, k) = (-(-8)/2, -(-2)/2) = (4, 1); r = √(16+1-8) = 3
