Question 1091909
<br>It took me several times looking at this problem before I could make sense out of it.  I think I understand now....<br>
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.<br>
(1) Find P(win) if the first dart is 4.<br>
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%<br>
(2) Find P(win) if the first dart is 3.<br>
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%<br>
(3) Find P(win) if the first dart is 2.<br>
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%<br>
(4) Find P(lose) if the first dart is 8.<br>
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%<br>