Becky Hamilton
New Member
Find the value of |A| if:
(1) |P(A)| = 4
(2) |B| = |A|+ 1 and |A×B| = 30
(3) |B| = |A|+ 2 and |P(B)|−|P(A)| = 24
(1) |P(A)| = 4
(2) |B| = |A|+ 1 and |A×B| = 30
(3) |B| = |A|+ 2 and |P(B)|−|P(A)| = 24
Please help!!
- Thread starter Becky Hamilton
- Start date
(1), how can probability be greater than 1?
Becky Hamilton
New Member
It is a set theory question
(1) |P(A)| = 4 <-- Cardinality of the power set is 4, which means we have 2^n = 4. |A| = 2
(2) |B| = |A|+ 1 and |A×B| = 30
|B| = 6 if |A| = 5 and |A x B| = 30
(3) |B| = |A|+ 2 and |P(B)|−|P(A)| = 24
Since |B| = |A|+ 2, we have: 2^(a + 2) - 2^a = 24
2^a(2^2 - 1) = 24
2^a(3) = 24
2^a = 8
|A |= 3
To check, we have |B| = |A| + 2 --> 3 + 2 = 5
So |P(B)| = 2^5 = 32
|P(A)| = 2^3 = 8
And 32 - 8 = 24
(2) |B| = |A|+ 1 and |A×B| = 30
|B| = 6 if |A| = 5 and |A x B| = 30
(3) |B| = |A|+ 2 and |P(B)|−|P(A)| = 24
Since |B| = |A|+ 2, we have: 2^(a + 2) - 2^a = 24
2^a(2^2 - 1) = 24
2^a(3) = 24
2^a = 8
|A |= 3
To check, we have |B| = |A| + 2 --> 3 + 2 = 5
So |P(B)| = 2^5 = 32
|P(A)| = 2^3 = 8
And 32 - 8 = 24