Problem Statement
The question tells us that the number of births for a whole year follows a uniform distribution. We are asked to find the following:
1. The probability that a child is born exact at the beginning of the 18th week.
2. Probability that a child is born between weeks 10 and 43.
3. P(x > 18 | x < 32)
Method
To evaluate the probability of a uniform distribution, we need to know a couple of things:
1. The probability that a child is born in weeks 1, 2, 3, and so on, up to week 53, is equal and the probability is:
[tex]P(1\text{ w}eek)=\frac{1}{52}[/tex]
2. The way to find the probability within a range of weeks is:
To find the probability between the two weeks, m and n, we simply find the area of the rectangle.
[tex]\begin{gathered} The\text{ length of the rectangle is:}Week\text{ n - W}eek\text{ m} \\ \text{width of the rectangle is: }\frac{1}{52} \\ \\ \text{Thus, the area is: }(n-m)\times\frac{1}{52} \end{gathered}[/tex]
With the above information, we can proceed to solve the question.
Implementation
1. The probability that a child is born exact at the beginning of the 18th week.
This means that m = 18 and n = 18. Thus, the probability that a child is born at the beginning of the 18th week is:
[tex]\begin{gathered} (n-m)\times\frac{1}{52} \\ P(\text{exactly 18 w}eeks)=(18-18)\times\frac{1}{52} \\ \\ P(\text{exactly 18 w}eeks)=0 \end{gathered}[/tex]
2. Probability that a child is born between weeks 10 and 43.
This means that m = 10 and n = 43. Thus the probability of being born between weeks 10 and 43 is:
[tex]\begin{gathered} P(\text{between 10 and 43)}=(43-10)\times\frac{1}{52} \\ \\ \therefore P(\text{between 10 and 43)}=\frac{33}{52} \end{gathered}[/tex]
3. P(x > 18 | x < 32)
This is a conditional probability as such, we can apply the Bayes theorem which states:
[tex]\begin{gathered} P(A|B)=\frac{P(A\cap B)}{P(B)} \\ A=\mleft\lbrace x>18\}\mright? \\ B=\mleft\lbrace x<32\}\mright? \\ \\ \therefore P(\mleft\lbrace x>18\mright\rbrace|\mleft\lbrace x<32\mright\rbrace)=\frac{P(\lbrace x>18\rbrace\cap\lbrace x<32\rbrace)}{P(\lbrace x<32\rbrace)} \\ \\ \lbrace x>18\rbrace\cap\lbrace x<32\rbrace\text{ tells us that x is within the range: 18 < x < 32. } \\ \therefore\P(\lbrace x>18\rbrace\cap\lbrace x<32\rbrace)=P(1818\rbrace|\lbrace x<32\rbrace)=\frac{14}{52}\div\frac{31}{52}=\frac{14}{52}\times\frac{52}{31} \\ \\ P(\lbrace x>18\rbrace|\lbrace x<32\rbrace)=\frac{14}{31}\approx0.4516 \end{gathered}[/tex]
