If P(A)=0.4, P(B)=0.5, and P(A∩B)=0.3, find a. P(A∪B), b. P(A∩B′), and c. P(A′∪B′).

"If P(A)=0.4, P(B)=0.5, and P(A∩B)=0.3, find a. P(A∪B),..." was answered by our community

Jessica Scott

Answered question

2021-11-20

P (A) = 0.4

P (B) = 0.5

, and

P (A \cap B) = 0.3

, find a.

P (A \cup B)

, b.

P (A \cap B^{'})

, and c.

P (A^{'} \cup B^{'})

Answer & Explanation

James Kilian

Beginner2021-11-21Added 20 answers

Explanation 1
Step 1
PROBABILITY RULES
Complement rule
$P (A^{'}) = P (\neg A) = 1 - P (A)$
General addition rule for any two events:
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
Step 2
SOLUTION
$P (A) = 0.4$
$P (B) = 0.5$
$P (A \cap B) = 0.3$
(a) Use the general addition rule for any two events:
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
$= 0.4 + 0.5 - 0.3$
$= 0.6$
Explanation 2
Step 1
Probability rules, used n the exercise:
Complement Rule:
$(A^{'}) = P (\neg A) = 1 - P (A)$
Addition Rule of general events:
$(A \cup B) = P (A) + P (B) - P (A \cap B)$
Rule of Complementary events B and B':
$P (A) = P (A \cap B) + P (A \cap B^{'})$
De Morgan's Rules
$P ({(A \cup B)}^{'}) = P (A^{'} \cap B^{'})$
$P ({(A \cap B)}^{'}) = P (A^{'} \cap B^{'})$
Step 2
Solution:
$P (A) = 0.4$
$P (B) = 0.5$
$P (A \cap B) = 0.3$
(a)
$\begin{matrix} (A d d i t i o n R u l e) \\ P (A \cup B) & = & P (A) - P (A \cap B) \end{matrix}$
$= 0.4 + 0.5 - 0.3$
$= 0.6$

Crom1970

Beginner2021-11-22Added 6 answers

Explanation 1
Step 3
(b) The event of A occurring but not B is the event A without all outcomes that occur for A and B have in common.
$P (A \cap B^{'}) = P (A) - P (A \cap B) = 0.4 - 0.3 = 0.1$
Explanation 2
(b)
$\begin{matrix} (R u l e o f C o m p l e m e n t a r y e v e n t s) \\ P (A \cap B^{'}) & = & P (A) - P (A \cap B) \end{matrix}$
$0.4 - 0.3$
$= 0.1$

Don Sumner

Skilled2021-11-23Added 184 answers

Explanation 1
Step 4
(c) The event of A not occurring or B not occurring is the event that A and B cannot occur together.
Use the complement rule:
$P (A^{'} \cup B^{'}) = P ({(A \cap B)}^{'}) = 1 - P (A \cap B) = 1 - 0.3 = 0.7$
Explanation 2
(c)
$\begin{matrix} (D e M o r g a n^{'} s R u l e (2)) \\ P (A^{'} \cup B^{'}) & = & P ({(A \cap B)}^{'}) \\ (C o m p l e m e n t R u l e) \\ = & 1 - P (A \cap B) \end{matrix}$
$= 1 - 0.3$
$= 0.7$

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