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Unformatted text preview: the drawing is without replacement . Then P (K on 1st, A on 2nd)= fav total = 4 52 ! 4 51 P (A on 3rd, K on 5th)= fav total = fav 52 ! 51 ! 50 ! 49 ! 48 For the fav, there are 5 slots to fill. The 3 rd slot can be filled in 4 ways, the 5 th slot in 4 ways, and then the other 3 slots in 50 ! 49 ! 48 ways. So P (A on 3rd, K on 5th)= 4 ! 4 ! 50 ! 49 ! 48 52 ! 51 ! 50 ! 49 ! 48 = 4 52 ! 4 51 The other 3 slots canceled out, leaving the same answer as P (K on 1st, A on 2nd) . Here is another way to solve this problem : P (K on 1st and A on 2nd)= P (K on 1st) P (AK on 1st) where P (K on 1st)= 4 52 P (AK on 1st)= 4 51 Example Draw without replacement from a box with 10 white and 5 black balls. To find the probability of W on the 1 st and 4 th draws (no information about the 2 nd and 3 rd ), take advantage of symmetry and switch to an easier problem: P (W on 1st and 4th)= P (W on 1st and 2nd)= 10 15 ! 9 14...
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 Spring '08
 Mendel

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