by Ben Ku Tue Aug 18, 2009 9:27 pm
sharok50's response only takes into account the probability of tossing exactly 5 heads in 6 tosses. However, it does not add in the probability of tossing 6 heads.
I look at the problem this way:
P(5 Heads) + P(6 Heads)
P(6 Heads) is easy. Each toss has a 0.6 chance of being heads, so P(6 Heads) = (0.6)^6.
P(5 Heads) is a bit more complicated, because there are different ways to get 5 heads. One combination of getting 5 Heads is HHHHHT. The probability of this combination is (0.6)(0.6)(0.6)(0.6)(0.6)(0.4) = (0.6)^5(0.4). [The 0.4 is there because that's the probability of getting tails.] Using the anagram method, there are 6!/5! = 6 different combinations of getting 5 Heads. So P(5 Heads) = (6)(0.6^5)(0.4) = (4)(0.6^6)
P(5 Heads) + P(6 Heads) = (4)(0.6^6) + (0.6^6) = 5(0.6^6)
Ben Ku
Instructor
ManhattanGMAT