A small college has 800 students, 10 percent of which are left-handed. Suppose we randomly select 8 students, because SRS = 8, and assume L = 3.

Also, assume we have to find the probability that exactly 3 out of the 8 students are left-handed, or the probability that L=3.

We use the binomial distribution for calculating this

^{n}C_{r} \cdot p^{r} (1-p)^{n-r}

p is the probability of successes which is 10% or 0.1, n = 8, r = 3. Plugging these in we get

^{8}C_{3}(0.10)^{3}(1-0.10)^{8-3}

= 56 (0.001)(0.9)^{5}

=0.033

So, the probability of getting 3 lefthanded students is 0.033.

I found an answer from ntrs.nasa.gov

20100028927.pdf

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I may be able to help you if you complete the question. Do you want to find the probability of the left handed students in the sample here or the mean and the standard deviation?

Here is a similar discussion based on the first few words in your statement which may help

https://www.qalaxia.com/viewDiscussion?messageId=60684c5ef4d26b0f37dbe1ba

I found an answer from www.khanacademy.org

**Binomial probability** example (video) | Khan Academy

We can use the **binomial distribution** to find the **probability** of getting a certain ...
**Binomial random variables** ... **Binomial** mean and standard deviation **formulas** ...
It may not be exact because this is experimental **data**, but it should be close. ...
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For more information, see **Binomial probability** example (video) | Khan Academy