Posted by Bersy on .
It is estimated that residents in 60% of households in town own a digital camera. If 10 homes are randomly selected, what is the probability that no more than 5 own digital cameras?

Math 
cassi,
60รท10=6*5=?

Math 
Reiny,
Unfortunately, this is not as simple as "cassi's" solution
This is a binomial distribution problem
We have to find the sum of the following cases
prob(none has one) + prob(1 of the 10 has one) + ... + prob (5 of 10 have one)
= C(10,0) (.6)^0 (.4)^10 > .0001049
+ C(10,1) (.6)^1 (.4)^9 > .001573
+ C(10,2) (.6)^2 (.4)^8 > .0106168
+ C(10,3) (.6)^3 (.4)^7 > .042467
+ C(10,4) (.6)^4 (.4)^6 > .1114767
+ C(10,5) (.6)^5 (.4)^5 > .200658
= appr .3244
(might be a good idea to check my calculations)