To solve this problem, we will use the binomial probability formula:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
n = total number of trials (10 individuals the carrier comes into contact with)
k = number of successful outcomes (individuals who contract tuberculosis)
p = probability of success in one trial (10%)
Now we need to calculate the probability that less than 6 individuals will contract tuberculosis:
P(X<6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)
P(X=0) = (10 choose 0) * 0.1^0 * 0.9^10 = 0.3487
P(X=1) = (10 choose 1) * 0.1^1 * 0.9^9 = 0.3874
P(X=2) = (10 choose 2) * 0.1^2 * 0.9^8 = 0.1937
P(X=3) = (10 choose 3) * 0.1^3 * 0.9^7 = 0.0574
P(X=4) = (10 choose 4) * 0.1^4 * 0.9^6 = 0.0106
P(X=5) = (10 choose 5) * 0.1^5 * 0.9^5 = 0.0011
Adding all these probabilities together:
P(X<6) = 0.3487 + 0.3874 + 0.1937 + 0.0574 + 0.0106 + 0.0011 = 0.9989
Therefore, the probability that less than 6 individuals will contract tuberculosis is approximately 0.9989, which is closest to 0.9999.
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
1 answer