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?
60รท10=6*5=?
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)
To find the probability that no more than 5 homes own digital cameras, we can use the binomial probability formula. The formula is:
P(x) = C(n, x) * p^x * (1 - p)^(n - x)
Where:
P(x) is the probability of getting exactly x successes
C(n, x) is the combination formula for choosing x items from a set of n items
p is the probability of success on a single trial
n is the total number of trials
In this case, we have n = 10 (number of homes selected) and p = 0.6 (probability of owning a digital camera).
To find the probability for each number from 0 to 5, we can substitute the values into the formula and sum them up.
P(0) = C(10, 0) * 0.6^0 * (1 - 0.6)^(10 - 0)
P(1) = C(10, 1) * 0.6^1 * (1 - 0.6)^(10 - 1)
P(2) = C(10, 2) * 0.6^2 * (1 - 0.6)^(10 - 2)
P(3) = C(10, 3) * 0.6^3 * (1 - 0.6)^(10 - 3)
P(4) = C(10, 4) * 0.6^4 * (1 - 0.6)^(10 - 4)
P(5) = C(10, 5) * 0.6^5 * (1 - 0.6)^(10 - 5)
Finally, we sum up the probabilities:
P(no more than 5) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5)
Let's calculate it step by step:
P(0) = C(10, 0) * 0.6^0 * (1 - 0.6)^(10 - 0)
= 1 * 1 * (0.4)^10
= 0.0060466176
P(1) = C(10, 1) * 0.6^1 * (1 - 0.6)^(10 - 1)
= 10 * 0.6 * (0.4)^9
= 0.0403107824
P(2) = C(10, 2) * 0.6^2 * (1 - 0.6)^(10 - 2)
= 45 * (0.6)^2 * (0.4)^8
= 0.120932352
P(3) = C(10, 3) * 0.6^3 * (1 - 0.6)^(10 - 3)
= 10 * (0.6)^3 * (0.4)^7
= 0.214990848
P(4) = C(10, 4) * 0.6^4 * (1 - 0.6)^(10 - 4)
= 210 * (0.6)^4 * (0.4)^6
= 0.250822656
P(5) = C(10, 5) * 0.6^5 * (1 - 0.6)^(10 - 5)
= 252 * (0.6)^5 * (0.4)^5
= 0.2006581248
Now, let's sum up the probabilities:
P(no more than 5) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5)
= 0.0060466176 + 0.0403107824 + 0.120932352 + 0.214990848 + 0.250822656 + 0.2006581248
= 0.8337613808
Therefore, the probability that no more than 5 homes own digital cameras is approximately 0.8338 (or 83.38%).