Question
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 a day, he meets 10 such individuals. Calculate the probability that 6 of these individuals will contract tuberculosis, we determine?
0.1
0.6251
0.9999
0.0001
0.9
0.1
0.6251
0.9999
0.0001
0.9
Answers
GPT-4o mini
The scenario described can be modeled by a binomial distribution, where we want to calculate the probability of exactly 6 out of 10 individuals contracting tuberculosis, given that the probability of any one individual contracting the disease is 10% (or 0.1).
The formula for the probability of getting exactly \( k \) successes (in this case, 6 individuals contracting the disease) in \( n \) trials (10 individuals) is given by the binomial probability formula:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
Here:
- \( n = 10 \) (the number of individuals)
- \( k = 6 \) (the number of individuals contracting TB)
- \( p = 0.1 \) (the probability of contracting TB)
1. Calculate \( \binom{n}{k} \):
\[
\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210
\]
2. Calculate \( p^k \):
\[
(0.1)^6 = 0.000001
\]
3. Calculate \( (1-p)^{n-k} \):
\[
(0.9)^{4} = 0.6561
\]
4. Putting it all together:
\[
P(X = 6) = 210 \times 0.000001 \times 0.6561
\]
\[
P(X = 6) = 210 \times 0.0000006561 = 0.000137781
\]
The calculated probability \( P(X = 6) \) is approximately \( 0.0001 \).
Thus, the correct answer is approximately:
\(\boxed{0.0001}\)
The formula for the probability of getting exactly \( k \) successes (in this case, 6 individuals contracting the disease) in \( n \) trials (10 individuals) is given by the binomial probability formula:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
Here:
- \( n = 10 \) (the number of individuals)
- \( k = 6 \) (the number of individuals contracting TB)
- \( p = 0.1 \) (the probability of contracting TB)
1. Calculate \( \binom{n}{k} \):
\[
\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210
\]
2. Calculate \( p^k \):
\[
(0.1)^6 = 0.000001
\]
3. Calculate \( (1-p)^{n-k} \):
\[
(0.9)^{4} = 0.6561
\]
4. Putting it all together:
\[
P(X = 6) = 210 \times 0.000001 \times 0.6561
\]
\[
P(X = 6) = 210 \times 0.0000006561 = 0.000137781
\]
The calculated probability \( P(X = 6) \) is approximately \( 0.0001 \).
Thus, the correct answer is approximately:
\(\boxed{0.0001}\)