# Homework Help: Biostats, Combination and Probability

1. Sep 23, 2008

### h6872

1. The problem statement, all variables and given/known data

A bite from a coastal taipan proves fatal 30% of the time. If three Queenslanders are bitten during 1 year, what is the probability that all 3 will die? That exactly 2 will die? That at most 1 will die?

3. The attempt at a solution

Assuming independence, then the probability of all three dying will simply be 0.3 x 0.3 x 0.3. For the second, if it had been 'at least 2 people', then the probability would be the probability of the two deaths times the probability of the one survival (0.3 x 0.3 x 0.7=0.063). However, the 'exactly' implies that a combination is required. For that, I did 3 choose 2 (= 3)and finally multiplied this by the probability that I would get for at least 2 deaths, which gave me 3 x 0.063 = 0.189.

Now, the solutions manual tells me that the probability for at most 1 dying is 1 - (P(all 3 die) + P(exactly 2 die))

Why is this? If I was looking at the probability of 1 death = 1 - (complement of P(1 death)). But if this was true, why would the event in which EXACTLY 2 die be taken into account? Why wouldn't it be the situation in which AT LEAST 2 dying occurred (0.3 x 0.3 x 0.7)? Or does this situation not take place in the sample space?