To solve these problems, we need to use the binomial distribution. The binomial distribution helps us calculate the probability of a given number of successes (or failures) out of a fixed number of trials. Here, the trials represent each child born, "success" is a child dying, and we have a fixed number of children (900).
Given data:
- Number of trials (children): [tex]\( n = 900 \)[/tex]
- Probability of a child dying (success): [tex]\( p = \frac{4}{1000} = 0.004 \)[/tex]
- Probability of a child not dying: [tex]\( q = 1 - p = 0.996 \)[/tex]
The binomial probability formula is given by:
[tex]\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot q^{n-k} \][/tex]
Where:
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, representing the number of ways to choose [tex]\( k \)[/tex] successes in [tex]\( n \)[/tex] trials.
- [tex]\( k \)[/tex] is the number of successes (in this case, the number of children who die).
### (a) Probability that none will die ([tex]\( k = 0 \)[/tex])
The probability that no child will die is given by:
[tex]\[ P(X = 0) = \binom{900}{0} \cdot (0.004)^0 \cdot (0.996)^{900} \][/tex]
From the results, this probability is:
[tex]\[ P(X = 0) = 0.02712717576551175 \][/tex]
### (b) Probability that at most 3 will die ([tex]\( k \leq 3 \)[/tex])
To find the probability that at most 3 children will die, we need to sum the probabilities for [tex]\( k = 0, 1, 2, \)[/tex] and [tex]\( 3 \)[/tex]:
[tex]\[ P(X \leq 3) = \sum_{k=0}^{3} \binom{900}{k} \cdot (0.004)^k \cdot (0.996)^{900-k} \][/tex]
From the results, this probability is:
[tex]\[ P(X \leq 3) = 0.5149606567308483 \][/tex]
### (c) Probability that at least 2 will die ([tex]\( k \geq 2 \)[/tex])
To find the probability that at least 2 children will die, we can use the complement rule. That is, subtract the probability of fewer than 2 children dying from 1:
[tex]\[ P(X \geq 2) = 1 - P(X < 2) \][/tex]
[tex]\[ P(X < 2) = P(X = 0) + P(X = 1) \][/tex]
[tex]\[ P(X \geq 2) = 1 - \left( P(X = 0) + P(X = 1) \right) \][/tex]
From the results:
[tex]\[ P(X \geq 2) = 0.8748227913470964 \][/tex]
### (d) Probability that between 3 and 6 will die (inclusive) ([tex]\( 3 \leq k \leq 6 \)[/tex])
To find the probability that between 3 and 6 children will die, we sum the probabilities for [tex]\( k = 3, 4, 5, \)[/tex] and [tex]\( 6 \)[/tex]:
[tex]\[ P(3 \leq X \leq 6) = \sum_{k=3}^{6} \binom{900}{k} \cdot (0.004)^k \cdot (0.996)^{900-k} \][/tex]
From the results, this probability is:
[tex]\[ P(3 \leq X \leq 6) = 0.6249447552364134 \][/tex]
In summary:
- (a) Probability that none will die: [tex]\( 0.02712717576551175 \)[/tex]
- (b) Probability that at most 3 will die: [tex]\( 0.5149606567308483 \)[/tex]
- (c) Probability that at least 2 will die: [tex]\( 0.8748227913470964 \)[/tex]
- (d) Probability that between 3 and 6 will die: [tex]\( 0.6249447552364134 \)[/tex]
