Question

1. given the universal set, set u = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.

-- determine the elements of the following subsets
-- draw a venn diagram of the subset
a) set b = the multiples of 4 =
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b) set f = # less than 5 =
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-- determine the probability
-- shade the required region
p(the multiples of 4) = 3/12
shade p(the multiples of 4)
p(# less than 5) = 4/12
shade p(# less than 5)
c) set p = prime #s = 2,3,5,7,11
p(prime #s) = 5/12
shade p(prime #s)
d) set k = even numbers =
p(even numbers) = 6/12
shade p(even numbers)

Explanation:

Step1: Determine elements of Set B (multiples of 4)

Multiples of 4 in \(U=\{1,2,3,4,5,6,7,8,9,10,11,12\}\) are \(4,8,12\). So \(n(B) = 3\).
The formula for probability \(P(B)=\frac{n(B)}{n(U)}\), where \(n(U)=12\).
So \(P(B)=\frac{3}{12}=\frac{1}{4}\).

Step2: Determine elements of Set F (\(\#\) less than 5)

Numbers less than 5 in \(U\) are \(1,2,3,4\). So \(n(F)=4\).
Using the probability formula \(P(F)=\frac{n(F)}{n(U)}\), with \(n(U) = 12\).
So \(P(F)=\frac{4}{12}=\frac{1}{3}\).

Step3: Determine elements of Set \(K\) (even numbers)

Even numbers in \(U\) are \(2,4,6,8,10,12\). So \(n(K)=6\).
Using the probability formula \(P(K)=\frac{n(K)}{n(U)}\), with \(n(U)=12\).
So \(P(K)=\frac{6}{12}=\frac{1}{2}\).

Step4: Determine elements of Set \(P\) (prime numbers)

Prime numbers in \(U\) are \(2,3,5,7,11\). So \(n(P)=5\).
Using the probability formula \(P(P)=\frac{n(P)}{n(U)}\), with \(n(U)=12\).
So \(P(P)=\frac{5}{12}\).

Answer:

a) \(P(\text{multiples of }4)=\frac{1}{4}\)
b) \(P(\#\text{ less than }5)=\frac{1}{3}\)
c) \(P(\text{prime }\#\text{'s})=\frac{5}{12}\)
d) \(P(\text{even numbers})=\frac{1}{2}\)