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 calculate the probability that no more than 5 homes own digital cameras out of 10 randomly selected homes, we need to use the binomial distribution formula.

The binomial distribution formula is:

P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)

Where:
- P(X = k) is the probability of having exactly k successes
- n is the total number of trials or homes selected (10 in this case)
- k is the number of successes (no more than 5 in this case)
- p is the probability of success on any given trial (60% or 0.6 in this case)
- (n choose k) is the number of combinations or ways to choose k successes out of n trials

To calculate the probability of no more than 5 homes owning digital cameras, we need to find the sum of probabilities for k = 0, 1, 2, 3, 4, and 5:

P(X <= 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

Let's calculate these probabilities step by step:

P(X = 0) = (10 choose 0) * (0.6^0) * (0.4^10)

(10 choose 0) represents the number of ways to choose 0 successes out of 10 trials, which is simply 1.

(0.6^0) is equal to 1 since any number raised to the power of 0 is 1.

(0.4^10) represents the probability of no successes in 10 trials. It is calculated by raising the probability of failure (1 - p) to the power of n, where n is the total number of trials.

Using a calculator, we can evaluate it to be approximately 0.00605.

P(X = 1) = (10 choose 1) * (0.6^1) * (0.4^9)
P(X = 2) = (10 choose 2) * (0.6^2) * (0.4^8)
P(X = 3) = (10 choose 3) * (0.6^3) * (0.4^7)
P(X = 4) = (10 choose 4) * (0.6^4) * (0.4^6)
P(X = 5) = (10 choose 5) * (0.6^5) * (0.4^5)

Using a calculator, we can calculate these probabilities as follows:
P(X = 1) ≈ 0.04032
P(X = 2) ≈ 0.12096
P(X = 3) ≈ 0.21504
P(X = 4) ≈ 0.25082
P(X = 5) ≈ 0.20066

Now, we can sum up all these probabilities:

P(X <= 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
≈ 0.00605 + 0.04032 + 0.12096 + 0.21504 + 0.25082 + 0.20066
≈ 0.83385

Therefore, the probability that no more than 5 homes own digital cameras out of 10 randomly selected homes is approximately 0.83385 or 83.38%.