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.
0.9
0.1
0.9437
0.9874
0.9999
To calculate the probability that less than 6 individuals will contract tuberculosis, we can use the binomial distribution formula:
P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
where X is the number of individuals who contract tuberculosis.
Given that the carrier has a 10% chance of passing on tuberculosis, the probability of an individual not contracting tuberculosis is 0.9.
Using the binomial distribution formula, we calculate:
P(X < 6) = ((10 choose 0) * 0.9^0 * 0.1^10) + ((10 choose 1) * 0.9^1 * 0.1^9) + ((10 choose 2) * 0.9^2 * 0.1^8) + ((10 choose 3) * 0.9^3 * 0.1^7) + ((10 choose 4) * 0.9^4 * 0.1^6) + ((10 choose 5) * 0.9^5 * 0.1^5)
Calculating each term individually:
P(X = 0) = (10 choose 0) * 0.9^0 * 0.1^10 = 1 * 1 * 0.0000000001 = 0.0000000001
P(X = 1) = (10 choose 1) * 0.9^1 * 0.1^9 = 10 * 0.9 * 0.000000001 = 0.000000009
P(X = 2) = (10 choose 2) * 0.9^2 * 0.1^8 = 45 * 0.81 * 0.00000001 = 0.000000036
P(X = 3) = (10 choose 3) * 0.9^3 * 0.1^7 = 120 * 0.729 * 0.0000001 = 0.000000876
P(X = 4) = (10 choose 4) * 0.9^4 * 0.1^6 = 210 * 0.6561 * 0.000001 = 0.000138915
P(X = 5) = (10 choose 5) * 0.9^5 * 0.1^5 = 252 * 0.59049 * 0.00001 = 0.001376385
Adding all the individual probabilities together:
P(X < 6) = 0.0000000001 + 0.000000009 + 0.000000036 + 0.000000876 + 0.000138915 + 0.001376385 = 0.0015152
Therefore, the probability that less than 6 individuals will contract tuberculosis is 0.0015152 or approximately 0.0015.