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? Pick one:
A) (0.6)^5
B) 2(0.6)^4
C) 3(0.6)^4
D) 4(0.6)^4(0.4) + (0.6)^5
E) 5(0.6)^4(0.4) + (0.6)^5
P(h)=0.6P(h)=0.6, so P(t)=0.4P(t)=0.4. We want to determine the probability of at least 4 heads in 5 tries.   At least 4 heads means 4 or 5. Let's calculate each one:   5 heads: P(h=5)=0.6^5;   4h and 1t: P(h=4)=5!/4!∗0.6^4∗0.4=5∗0.6^4∗0.4, multiplying by 5 as 4h and 1t may occur in 5 different ways:   hhhht   hhhth   hhthh   hthhh   thhhh   So, P(h≥4)=0.6^5+5∗0.6^4∗0.4   Answer: 5(0.6)^4(0.4) + (0.6)^5.
<result>E</result>
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?