A carrier of tuberculosis has a 10% chance of passing the disease on to anyone with whom he comes into close contact who has had no prior exposure. During the course of a day, he comes into contact with 10 such individuals. Calculate the probability that less than 6 of these individuals will contract tuberculosis.
To calculate the probability that less than 6 of these individuals will contract tuberculosis, we can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where:
n = number of trials (10 individuals)
k = number of successes (less than 6)
p = probability of success (10% or 0.1)
Now, we need to calculate the probability for k = 0, 1, 2, 3, 4, and 5 and sum them up:
P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
= (10 choose 0) * 0.1^0 * 0.9^10 + (10 choose 1) * 0.1^1 * 0.9^9 + (10 choose 2) * 0.1^2 * 0.9^8 + (10 choose 3) * 0.1^3 * 0.9^7 + (10 choose 4) * 0.1^4 * 0.9^6 + (10 choose 5) * 0.1^5 * 0.9^5
= 0.3487
So, the probability that less than 6 of these individuals will contract tuberculosis is approximately 0.3487 or 34.87%.