# Thread: Simple solutions via complements

1. ## Simple solutions via complements

Hi,

I've got some questions about a fairly straightforward probability question. I've never taken a probability course but I think these are solvable via basic probability. However, I'd like to know if there is a systematic approach that one would generally use if the 'trick' of using complements of sample spaces was not available.

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Workers are assigned to different tasks each month in a company.
55% of employees have less than or equal to 2 years of experience
32% have between three and five years
13% have more than five

Assume that teams of 3 employees are randomly formed. Dave is in such a team.

a) What is the prob. that (at least) one of Dave's teammates has
1. two or fewer years of experience?
2. more than five years of experience?

b) What is the prob. considering both of Dave's two teammates,
1. both have more than two years
2. exactly one has more than five years of experience
3. at least one has more than five years of experience?

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For a1., I would just ask what is the prob. that neither of them have less than 2 yrs, which is (1-0.55)^2, so the prob that at least one has 2 years is 1 - 0.45^2.

a2. Same thing, P(at least one >5) = 1 - P(none of them 5) = 1 - (1 - 0.13)^2.
If they mean exactly one, wouldn't that be b2.?

b1. I'd say (0.45)^2
2. Conditional probability here? I'd say (0.13)(0.87) * 2 since there are two teammates.
3. Seems the same as a2.

Can anyone confirm that these answers are correct?

Cheers,

Michael

2. In a) Are you considering only one teammate? "One" and "at least one" are two different things. =\

If so, the answers would be directly 0.55 for 1. and 0.13 for 2.