Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Biostats, Combination and Probability

  1. Sep 23, 2008 #1
    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?

    Please help me sort out this logic... I'm completely confused!
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted