Certainly! Let's analyze the given problem step-by-step to find the required probabilities.
We are given:
- The probability that a child is infected with cholera (p) = 0.20.
- The number of children selected at random (n) = 11.
- Let [tex]\( X \)[/tex] be the number of infected children among the 11 selected.
Since [tex]\( X \)[/tex] follows a binomial distribution [tex]\( \text{B}(n, p) \)[/tex], the probability mass function (PMF) and the cumulative distribution function (CDF) of a binomial random variable help in determining the various probabilities.
1. Probability that [tex]\( X \)[/tex] is at most 2:
To find [tex]\( P(X \leq 2) \)[/tex]:
[tex]\[ P(X \leq 2) = 0.6174015487999999 \][/tex]
2. Probability that [tex]\( X \)[/tex] is equal to 4:
To find [tex]\( P(X = 4) \)[/tex]:
[tex]\[ P(X = 4) = 0.11072962559999995 \][/tex]
3. Probability that [tex]\( X \)[/tex] is at least 4:
To find [tex]\( P(X \geq 4) \)[/tex]:
[tex]\[ P(X \geq 4) = 1 - P(X \leq 3) = 0.16113920000000004 \][/tex]
4. Probability that [tex]\( X \)[/tex] is greater than 7:
To find [tex]\( P(X > 7) \)[/tex]:
[tex]\[ P(X > 7) = 1 - P(X \leq 7) = 0.00023521280000005085 \][/tex]
5. Probability that [tex]\( X \)[/tex] is less than 9:
To find [tex]\( P(X < 9) \)[/tex]:
[tex]\[ P(X < 9) = P(X \leq 8) = 0.999981056 \][/tex]
Thus, the final probabilities are:
- [tex]\( P(X \leq 2) = 0.6174015487999999 \)[/tex]
- [tex]\( P(X = 4) = 0.11072962559999995 \)[/tex]
- [tex]\( P(X \geq 4) = 0.16113920000000004 \)[/tex]
- [tex]\( P(X > 7) = 0.00023521280000005085 \)[/tex]
- [tex]\( P(X < 9) = 0.999981056 \)[/tex]
These results align with the calculations for a binomial distribution with the given parameters.
