Question 1091909: Two darts are thrown at the board below, if you get the same number twice, the second dart is removed and re-thrown. If you get 10 points or more, you win a prize.
5324
6128
5137
2122
1) If a 4 was scored with the first dart, find P(Win)
2)If a 3 was scored with the first dart, find P(Win)
3)If a 2 was scored with the first dart, find P(Win)
4)If an 8 was scored with the first dart, find P(Lose)
Answer as a percentage- round to 1 d.p
Can someone please help me?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
It took me several times looking at this problem before I could make sense out of it. I think I understand now....
The dart board is a 4x4 array of squares with the numbers you show in the squares. You throw two darts; if the second one gets the same number as the first, you throw the second dart again. You win if the sum of the two numbers is 10 or more.
(1) Find P(win) if the first dart is 4.
There is only one 4 on the board, so there are 15 squares you can hit with the second dart. Of those 15, 3 have numbers large enough (6 or greater) that will give a combined score of 10 or more. So P(win) = 3/15 = 1/5 = 20%
(2) Find P(win) if the first dart is 3.
There are two 3's on the board, so there are 14 squares you can hit with the second dart. Of those 14, 2 have numbers large enough (7 or greater) that will give a combined score of 10 or more. So P(win) = 2/14 = 1/7 = 14.3%
(3) Find P(win) if the first dart is 2.
There are five 2's on the board, so there are 11 squares you can hit with the second dart. Of those 11, only 1 has a number large enough (8 or greater) that will give a combined score of 10 or more. So P(win) = 1/11 = 9.1%
(4) Find P(lose) if the first dart is 8.
There is only one 8 on the board, so there are 15 squares you can hit with the second dart. This time you want to lose, meaning you want a sum less than 10; of the 15 squares you can hit, 3 have a number small enough (it must be 1) that will give a combined score of less than 10. So P(lose) = 3/15 = 1/5 = 20%
|
|
|
