Question 373747:  OK...so the first part of this question asked to figure out for EXACTLY three people. I got an answer of 0.1691. The second half of the question I am stumped with. Here is the second question:
 
Suppose that the probability that a new medication will cause a bad side effect is 0.03.  If this medication is given to 150 people, what is the probability that more than three of them will experience a bad side effect?  Place your answer, rounded to 4 decimal places, in the blank.  
 
Thanks! 
 Answer by jim_thompson5910(35256)      (Show Source): 
You can  put this solution on YOUR website! 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 jim_thompson5910@hotmail.com
 
 
Also, feel free to check out my tutoring website
 
 
Jim 
  | 
 
  
 
 |   
 
 |