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

We want at least 4 heads. So, there are two scenarios that work:

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?)

Sorry about the 5th choice..just too lazy to type..will make sure include all choices from now onwards!!