Biostats, Combination and Probability

[h6872]

Homework Statement

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?

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!

 

[anathema]

Firstly, please note that this is a science forum and not a mathematics forum. However, I will do my best to help you understand the logic behind the solution to this problem.

The key concept to understand in this problem is the difference between "at least" and "exactly". When we say "at least 2 will die", it includes all cases where 2 or more people die. However, when we say "exactly 2 will die", it only includes the specific case where exactly 2 people die and not any other combinations.

Now, let's look at the solution provided in the manual. The probability for at most 1 dying is 1 - (P(all 3 die) + P(exactly 2 die)). This is because the event of "at most 1 dying" includes all cases where 0 or 1 people die. So, we take the total probability (1) and subtract the probabilities of the specific cases where all 3 die and exactly 2 die.

To clarify further, let's look at the probabilities for each case separately:

1. All 3 die: 0.3 x 0.3 x 0.3 = 0.027
2. Exactly 2 die: (3 choose 2) x (0.3 x 0.3 x 0.7) = 0.189

So, the total probability for "at most 1 dying" is 1 - (0.027 + 0.189) = 0.784.

I hope this explanation helps you understand the logic behind the solution. Remember, it's important to pay attention to the wording of the problem and to understand the difference between "at least" and "exactly" in probability questions.

 

[baba89]

I would say that the solution manual is correct. The probability of at most 1 dying is the complement of the probability of all 3 dying or exactly 2 dying. This is because the event of exactly 2 dying is included in the event of at most 1 dying. For example, if we have 3 people and we want to know the probability of at most 1 dying, this could happen in three ways: 1 person dies, 2 people die, or all 3 people die. The event of exactly 2 dying is included in the event of 2 people dying. Therefore, to calculate the probability of at most 1 dying, we need to subtract the probability of all 3 dying and exactly 2 dying from 1.

In this case, the probability of exactly 2 dying is 0.189, as calculated in the attempt at a solution. The probability of all 3 dying is 0.3 x 0.3 x 0.3 = 0.027. Therefore, the probability of at most 1 dying is 1 - (0.189 + 0.027) = 0.784.

It is important to consider all possible outcomes and events when calculating probabilities, even if they seem redundant. In this case, the event of exactly 2 dying is not the same as the event of at least 2 dying, and therefore it must be taken into account separately.