We want at least 4 heads. So, there are two scenarios that work:moneyman wrote:For one toss of a certain coin, the probability that the outcome is heads in 0.6. If the coin is tossed 5 times, which of the following is the probability that the outcome will be heads atleast 4 times??
(0.6)^5
2(0.6)^4
3(0.6)^4 (0.4)
5(0.6)^4 (0.4) + (0.6)^5
Ans is D
4H/1T and 5H.
Since we want 4 heads OR 5 heads, we ADD the individual probabilities.
5 heads is easier, let's start there:
(.6)(.6)(.6)(.6)(.6) = (.6)^5.
Hey, let's look at the answer choices, our best buddies on test day. Only one of the choices is something + (.6)^5 - a confident test taker will pick that choice and move on!
If this were question 37 and we had 5 minutes left, we'd actually solve the other part:
the probability of getting 4h and 1t is more complicated, since we also worry about order. We could have HHHHT, HHHTH, HHTHH, HTHHH, THHHH - 5 different arrangements.
So, the proability of 4H and 1T is:
5 * (.6)^4 * (.4)^1 = 5(0.6)^4 (0.4)
Putting the two parts together, we get:
(5 * (.6)^4 * (.4)) + (.6)^5
(btw, what happened to the 5th choice?)
