Question 373747
There are two ways to do this....



1) The hard way: Find the probability P(X > 3). This is equivalent to saying P(X = 4)+P(X = 5) + ....P(X = 149)+P(X = 150)


That's a lot of work. It's certainly doable, but let's see if there's an easier way.



2) The easy way: First find the probability P(X <= 3) and then use the idea that EVERY individual probability MUST add to 1. This idea is crucial and is used throughout stats.



So find P(X = 0) (ie the probability that none suffer side effects) 


then find P(X = 1) (ie the probability that one person suffers side effects) 


then find P(X = 2) (ie the probability that two people suffer side effects) 


and finally, find P(X = 3) (ie the probability that three people suffer side effects). You've already done this.



Already, we see that this is an easier task because there are only 4 items to sum (compared to 147 items)



So because P(X > 3) = 1-P(X <= 3), and P(X <= 3) = P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3), this is equivalent to saying



P(X > 3) = 1 - (P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3))



or



P(X > 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)



Hopefully, this is enough to get you going. If not let me know.



As a check, you can use the binomialcdf function found on many TI calculators. 



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=Algebra%20Help">jim_thompson5910@hotmail.com</a>


Also, feel free to check out my <a href="http://www.freewebs.com/jimthompson5910/home.html">tutoring website</a>


Jim