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April 18, 2015

Posted by **Shane** on Monday, March 22, 2010 at 8:44pm.

(a) How many different outcomes are possible?

(b) How many different outcomes have exactly 9 heads?

(c) How many different outcomes have at least 2 heads?

(d) How many different outcomes have at most 10 heads?

- Probabilities (Math) -
**Reiny**, Monday, March 22, 2010 at 10:05pma) each turn could be 2 ways, so 2^14 = 16384

b) this is the same as asking, "in how many ways can we arrange 9 H's and 5 T's, the H's and T's are indistinguishable.

number of ways = 14!/(9!5!) = 2002

c)at least two heads implies we don't want 0 heads, or 1 head, let's find those two

0 heads ---> 1 way

1 head -----> 14!/13! = 14

so at least 2 heads = 16384 - 14 - 1 = 16369

d) so we don't want 11 heads, 12 heads, 13 heads and 14 heads

which are 14!/(11!3!) + 14!/(12!2!) + 14!/(13!1!) + 1

= 364 + 91 + 14 + 1 = 470

so at most 10 heads = 16384 - 470 = 15914

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