Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:
In this probability problem, we first need to split up the successful cases into three batches:
a) Flipping zero tails
b) Flipping exactly one tails
c) Flipping exactly two tails
a) Flipping zero tails is the same as flipping five heads in a row (HHHHH). Since the probability of flipping one heads is 0.6, and the flips are independent, we get (0.6)5. Unfortunately, this term is common to every answer choice, perhaps not surprisingly.
b) Flipping exactly one tails is the same as flipping four heads and one tails. These can come in any of five orders, depending on the position of the “odd man out”:
Each of these sub-cases has a probability of (0.6)4(0.4). Since we have five separate sub-cases, any of which counts as success, we add up the probabilities to get 5(0.6)4(0.4). There is only one answer choice that fits: we could stop here.
For the sake of completeness, let’s do the last part.
c) Flipping exactly two tails is the same as flipping three heads and two tails in any order. To count the arrangements, we can either write them all out or use a combinatorics approach, such as the anagram method. Using the latter, we get 5!/(3!2!) = 10 arrangements. This 10 becomes the coefficient on the probability of one particular arrangement (such as HHTHT). We get 10(0.6)3(0.4)2 as the last term.
The correct answer is A.