Prob. Solving

[Changoo]

36 of 80 teachers at a local school are certified in CPR. In 180 days of school, assuming none of the teachers were ever absent, about how many days is it that the teacher on bus duty will likely be certified in CPR?

Can someone give some idea of how to solve this problem.

I notice that they want the probability that "the teacher on bus duty will likely be certified in CPR". So it is the probability of one teacher.


I did 180/36 =5 days. But I don't think that is write. I am currently learning discrete probability distributions.

I hope someone can help. Thanks

[vsage]

Let me ask you something. How many times (on average) will a certain teacher have to do bus duty? What is the ratio of teachers who know CPR to teachers who don't? What's the probability that "certain teacher" in question knows CPR?

[DivGradCurl]

It's just a guess...


1 day: \left(\frac{36}{80}\right)

2 days in a row: \left(\frac{36}{80}\right)^2

3 days in a row: \left(\frac{36}{80}\right)^3

Then,

180 days in a row: \left(\frac{36}{80}\right)^{180}

Again, I'm not sure this is correct.

[HallsofIvy]

It's just a guess...


1 day: \left(\frac{36}{80}\right)

2 days in a row: \left(\frac{36}{80}\right)^2

3 days in a row: \left(\frac{36}{80}\right)^3

Then,

180 days in a row: \left(\frac{36}{80}\right)^{180}

Again, I'm not sure this is correct.

What question is that supposed to be the answer to? The question originally asked said nothing about "days in a row".

[vsage]

You're thinking more than you have to. Answer my original questions and I think the answer will jump out you.

[DivGradCurl]

That's why I said I wasn't sure. :rofl: Thanks for pointing it out.

How about:

\left(\frac{36}{80}\right)\cdot 180 = 81

[vsage]

That's why I said I wasn't sure. :rofl: Thanks for pointing it out.

How about:

\left(\frac{36}{80}\right)\cdot 180 = 81

I believe so :)