In a reliability study of a power station, Event A is defined as an inverter system malfun… — GRE General Test

In a reliability study of a power station, Event $A$ is defined as an inverter system malfunction and Event $B$ is defined as a battery backup failure during peak hours. The probability that at least one of these two malfunctions occurs is $0.80$ , and the probability that an inverter system malfunction occurs without a battery backup failure is $0.35$ . If Event $A$ and Event $B$ are independent events, what is the probability of an inverter system malfunction, $P(A)$ ?

A
$\frac{7}{20}$
B
$\frac{9}{20}$
$\frac{7}{11}$ Answer
D
$\frac{11}{20}$
E
$\frac{4}{5}$

Answer

The probability of an inverter system malfunction,

P(A)

, is

\frac{7}{11}

The probability of the union of two events can be decomposed as

P(A \cup B) = P(A \cap B^c) + P(B)

. Given

P(A \cup B) = 0.80

and

P(A \cap B^c) = 0.35

, solving yields

P(B) = 0.45

. Because Event

A

and Event

B

are independent,

P(A \cap B) = P(A)P(B) = 0.45 P(A)

. Since

P(A \cap B^c) = P(A) - P(A \cap B)

, we substitute to get

0.35 = P(A) - 0.45 P(A) = 0.55 P(A)

. Solving for

P(A)

gives

P(A) = \frac{0.35}{0.55} = \frac{7}{11}

Step-by-Step Solution

Relate the union probability to the difference event

A \cap B^c

and event

B

P(A \cup B) = P(A \cap B^c) + P(B)

The union of two events

A

and

B

can be partitioned into the region where only

A

occurs (

A \cap B^c

) and all outcomes in

B

Calculate

P(B)

using the given values.

P(B) = P(A \cup B) - P(A \cap B^c) = 0.80 - 0.35 = 0.45 = \frac{9}{20}

Subtracting

P(A \cap B^c)

from

P(A \cup B)

yields

P(B)

Apply the definition of independent events to express

P(A \cap B)

in terms of

P(A)

P(A \cap B) = P(A) \cdot P(B) = 0.45 P(A)

For independent events, the joint probability is the product of their individual probabilities.

Set up and solve the equation for

P(A)

P(A \cap B^c) = P(A) - P(A \cap B) \implies 0.35 = P(A) - 0.45 P(A) = 0.55 P(A) \implies P(A) = \frac{0.35}{0.55} = \frac{7}{11}

Dividing

0.35

0.55

yields the exact value of

P(A)

Key Concept

Probability rules for independent events and set operations on sample spaces

Estimated Time:2m 0s

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