In a sample space of a random experiment, A and B are independent events such that P(A) = … — GRE General Test

In a sample space of a random experiment, $A$ and $B$ are independent events such that $P(A) = 0.35$ and $P(A \cup B) = 0.74$ . Event $C$ is mutually exclusive with Event $A$ . If the conditional probability $P(C \mid B) = 0.20$ , what is the probability that Event $B$ occurs, but neither Event $A$ nor Event $C$ occurs?

A
0.19
0.27Cevap
C
0.39
D
0.48
E
0.51

Cevap

0.27

To find the probability that Event

B

occurs without

A

C

, we must isolate the region of

B

that does not overlap with

A

C

. Since

A

and

B

are independent,

P(A \cup B) = P(A) + P(B) - P(A)P(B)

, which allows us to solve for

P(B) = 0.60

and

P(A \cap B) = 0.21

. Next, using the conditional probability

P(C \mid B) = 0.20

, we find

P(B \cap C) = 0.20 \times 0.60 = 0.12

. Because

A

and

C

are mutually exclusive, the intersections

A \cap B

and

B \cap C

do not overlap. Subtracting both intersection probabilities from

P(B)

yields

0.60 - 0.21 - 0.12 = 0.27

Adım Adım Çözüm

Calculate the probability of Event

B

P(B)

, using the independence of

A

and

B

P(B) = 0.60

Since

A

and

B

are independent,

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

. Substituting into the union formula

P(A \cup B) = P(A) + P(B) - P(A)P(B)

gives

0.74 = 0.35 + P(B)(1 - 0.35)

, so

0.39 = 0.65 P(B)

, yielding

P(B) = 0.60

Find the joint probability

P(A \cap B)

P(A \cap B) = 0.21

By independence,

P(A \cap B) = P(A) \times P(B) = 0.35 \times 0.60 = 0.21

Calculate the joint probability

P(B \cap C)

using the conditional probability formula.

P(B \cap C) = 0.12

From the definition of conditional probability,

P(C \mid B) = \frac{P(B \cap C)}{P(B)}

, so

P(B \cap C) = P(C \mid B) \cdot P(B) = 0.20 \times 0.60 = 0.12

Determine the probability that

B

occurs but neither

A

nor

C

occurs.

P(B \cap A^c \cap C^c) = 0.27

Since

A

and

C

are mutually exclusive, events

(A \cap B)

and

(B \cap C)

are disjoint subsets of

B

. Therefore,

P(B \cap A^c \cap C^c) = P(B) - P(A \cap B) - P(B \cap C) = 0.60 - 0.21 - 0.12 = 0.27

Anahtar Kavram

Probability Rules for Independent, Dependent, and Mutually Exclusive Events

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