2. It is assumed that 20% of the children in a locality are infected with cholera. Suppose that eleven children are selected at random. Let [tex]$ X $[/tex] equal the number of children that are infected. Find:

- [tex]$ P(X \leq 2) $[/tex]
- [tex]$ P(X = 4) $[/tex]
- [tex]$ P(X \geq 4) $[/tex]
- [tex]$ P(X \ \textgreater \ 7) $[/tex]
- [tex]$ P(X \ \textless \ 9) $[/tex]

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.