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100 Students want to enter a university. Each student has a 4% chance of entering (they all enter independently) What is the chance of atleast 3 entering the university?

The answer to this question is 0.7678

I just dont know how to get to this. I tried doing this:

Each student has a 96% of not entering, the chances of 3 students not entering therefore should be (0.96)^3. Now if i want the chances of atleast 3 entering i do 1 - (0.96)^3 = 0.115

Can someone explain me what i did wrong?

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Use the binomial distribution. Let $X$ be the number of students entering. Then $X\sim B(100,0.04)$.

$$P(X\geq3)=1-P(X\leq2)=1-P(X=0)-P(X=1)-P(X=2)$$

We have $$P(X=0)=0.96^{100}=0.0169$$ $$P(X=1)=100\times0.96^{99}\times0.04=0.0703$$ $$P(X=2)={{100}\choose{2}}0.96^{98}\times0.04^2=0.1450$$ Thus $P(X\geq3)=1-0.0169-0.0703-0.1450=0.7678$.

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  • $\begingroup$ Probably a really stupid question but, how did you know you had to use binomial distribution? $\endgroup$ – D.g Aug 19 '18 at 18:53
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    $\begingroup$ @D.g Not a stupid question at all. There are a fixed number of people (100) and the probability of success is the same each time (0.04). When you have a fixed number of trials and a constant probability of success, it's usually appropriate to use the binomial distribution. $\endgroup$ – A. Goodier Aug 19 '18 at 18:56
  • $\begingroup$ omg yes! Because it says independent i just kept it at a 100 people everytime, stupid of me. Thank alot for helping me. $\endgroup$ – D.g Aug 19 '18 at 18:57

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