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by Stuart@KaplanGMAT » Sun May 31, 2009 9:00 pm
We want the probability of at least 4 heads. Therefore, there are 2 results that make us happy: 4 heads or 5 heads. Since we want 4 heads OR 5 heads, we calculation the probability of each occuring and ADD those probabilities together.

Ignoring the order of flips, the chance of getting 4 heads and 1 tail is:

(.6)(.6)(.6)(.6)(.4)

Since the tails could occur in 5 different spots (1st, 2nd, 3rd, 4th or 5th), there are 5 different sequences that give us 4H and 1T. Therefore, the probability of getting exactly 4 heads is:

5 * (.6)^4 * (.4)

There's only 1 way to get 5 heads: HHHHH. The probability of getting 5 heads is:

(.6)(.6)(.6)(.6)(.6) = (.6)^5

Now we add those two results together to get our final answer:

5(.6)^4(.4) + (.6)^5
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by ghacker » Wed Jun 03, 2009 9:01 pm
The question doesn't has that the order is important it says that you have to get only 4 or more heads

so you can get 4 heads or you can get 5 heads out of 5 trials

how many ways are there to get 4 heads ?

you can write it down : HHHHT , but this T can be any where and all H are the same so HHHHT can be arranged in 5! ways but inside the 5! ways there are similar arrangements how many 4!

so total no of ways = 5!/4! = 5

hence the answer is 5.(0.6)^4(0.2)+(0.6)^5

We can look at the H-T arrangement in a different way

_H_H_H_H_ T can fill the _ so only 5 ways
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