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 the course of a day, he comes into contact with 10 such individuals. Calculate the probability that 6 of these individuals will contract tuberculosis, we determine?
Answers
GPT-4o mini
To solve this problem, we can model the situation using the binomial probability formula.
The binomial probability formula is given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where:
- \( n \) is the number of trials (in our case, the number of individuals contacted, which is 10),
- \( k \) is the number of successes (the number of individuals who contract tuberculosis; here, we want \( k = 6 \)),
- \( p \) is the probability of success on an individual trial (the probability that an individual contracts tuberculosis, which is 0.10),
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).
Given our parameters:
- \( n = 10 \)
- \( k = 6 \)
- \( p = 0.10 \)
We need to calculate \( P(X = 6) \):
1. **Calculate the binomial coefficient** \( \binom{10}{6} \):
\[
\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210
\]
2. **Calculate \( p^k = (0.10)^6 \)**:
\[
(0.10)^6 = 0.000001
\]
3. **Calculate \( (1-p)^{n-k} = (0.90)^{4} \)**:
\[
(0.90)^{4} = 0.6561
\]
4. **Putting it all together**:
Now, plug everything into the binomial probability formula:
\[
P(X = 6) = \binom{10}{6} (0.10)^{6} (0.90)^{4}
\]
\[
P(X = 6) = 210 \times 0.000001 \times 0.6561
\]
Calculating that gives:
\[
P(X = 6) = 210 \times 0.0000006561 = 0.000137781
\]
Thus, the probability that exactly 6 individuals will contract tuberculosis is approximately:
\[
\boxed{0.0001378}
\]
The binomial probability formula is given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where:
- \( n \) is the number of trials (in our case, the number of individuals contacted, which is 10),
- \( k \) is the number of successes (the number of individuals who contract tuberculosis; here, we want \( k = 6 \)),
- \( p \) is the probability of success on an individual trial (the probability that an individual contracts tuberculosis, which is 0.10),
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).
Given our parameters:
- \( n = 10 \)
- \( k = 6 \)
- \( p = 0.10 \)
We need to calculate \( P(X = 6) \):
1. **Calculate the binomial coefficient** \( \binom{10}{6} \):
\[
\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210
\]
2. **Calculate \( p^k = (0.10)^6 \)**:
\[
(0.10)^6 = 0.000001
\]
3. **Calculate \( (1-p)^{n-k} = (0.90)^{4} \)**:
\[
(0.90)^{4} = 0.6561
\]
4. **Putting it all together**:
Now, plug everything into the binomial probability formula:
\[
P(X = 6) = \binom{10}{6} (0.10)^{6} (0.90)^{4}
\]
\[
P(X = 6) = 210 \times 0.000001 \times 0.6561
\]
Calculating that gives:
\[
P(X = 6) = 210 \times 0.0000006561 = 0.000137781
\]
Thus, the probability that exactly 6 individuals will contract tuberculosis is approximately:
\[
\boxed{0.0001378}
\]