For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads 'at least' 4 times?

(A) (0.6)^5

(B) 2(0.6)^4

(C) 3[(0.6)^4](0.4)

(D) 4[((0.6)^4)(0.4)] + (0.6)^5

(E) 5[((0.6)^4)(0.4)] + (0.6)^5

Answer is (E), but I have no idea how it was obtained?

Hi Liddar,

I used your screen shot to clean up the formatting of your question, and am posting an edited version of Stacey's original response to your question.

There are two possible ways of getting at least four heads: EITHER four heads and one tail, OR five heads. So the probability of getting at least four heads is just (the probability of getting four heads and 1 tail) PLUS (the probability of getting five heads). Remember that for mutually exclusive events, "OR" means "ADD" in probability.

There is only one way of getting 5 heads - you must get 5 heads in a row, or HHHHH. The probability of this happening is (0.6)(0.6)(0.6)(0.6)(0.6)=(0.6)^5.

There are 5 different ways of getting 4 heads and 1 tail. You could get THHHH, or HTHHH, or HHTHH, or HHHTH, or HHHHT. The probability of getting THHHH is (0.4)(0.6)(0.6)(0.6)(0.6)=((0.6)^4)(0.4). The probability of getting HTHHH is (0.6)(0.4)(0.6)(0.6)(0.6), which also equals ((0.6)^4)(0.4). By similar reasoning, we can establish that the probability of getting any particular one of the 5 possible configurations is just ((0.6)^4)(0.4). Therefore, the probability of getting 4 heads and 1 tail is just 5[((0.6)^4)(0.4)]

Adding together (the probability of getting 4 heads and 1 tail) and (the probability of getting 5 heads), we get:
5[((0.6)^4)(0.4)] + ((0.6)^5)

The answer is E.