Need Help: Counting methods & Probability questions

harmeet wrote:The questions are -

1. You are taking a test that has 10 questions, and the answer to each question will either be "all", "some", or "none." If you leave none blank, in how many ways can you answer the questions?

a. 3000
b. 19,683
c. 59,049
d. 30

There are three ways to answer each of the question. Total no of ways = 3*3*...*3 (10 times) = 3^10 = 59049 (3^10 has to end with a 9)


2. A person can purchase a $50,000 flight insurance policy for an international flight at a cost of $50. Past history indicates that the probability of a person dying on an international flight with this particular airline is 0.0012. Determine the insurance company's expected gain per policy.
[Hint]

a. $29.88
b. $19.98
c. $39.94
d. $49.94

Expected cost = Expected probability of a person dying on an international flight* Cover offered
=0.0012* 50000
=60 dollars

In which case the insurance company will suffer a loss of 10 dollars per policy which is nonsensical
[but If we take 0.0012 is a percentage figure in which case the cost becomes 0.6 dollars and the gain is 49.4]