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?
To find the probability that no more than 5 homes own digital cameras, we need to calculate the probability of selecting 0, 1, 2, 3, 4, or 5 homes out of the 10 randomly selected homes.
The probability of selecting x homes owning digital cameras out of n homes can be calculated using the binomial probability formula:
P(x) = C(n,x) * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of getting x successes (homes owning digital cameras)
C(n,x) is the number of combinations of n items taken x at a time (calculated as n! / (x!(n-x)!)
p is the probability of getting a success (probability of owning a digital camera)
1-p is the probability of getting a failure (probability of not owning a digital camera)
n is the total number of trials (in this case, 10 randomly selected homes)
Given that 60% of households own a digital camera, the probability of a home owning a digital camera (p) is 0.6.
Now let's calculate the probability for each case:
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)
To find the probability that no more than 5 homes own digital cameras, we sum all the probabilities calculated above:
P(no more than 5) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5)
Let's calculate this now.