Posted by **Bersy** on Monday, January 28, 2013 at 1:05pm.

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**, Monday, January 28, 2013 at 1:08pm
60÷10=6*5=?

- Math -
**Reiny**, Monday, January 28, 2013 at 1:31pm
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)

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