A quality control analyst evaluates a manufacturing process in which two specific types of… — GRE General Test

A quality control analyst evaluates a manufacturing process in which two specific types of flaws, Flaw $X$ and Flaw $Y$ , can occur on produced glass panels. The probability that a randomly selected panel has Flaw $X$ is $P(X) = 0.20$ , and the probability that it has Flaw $Y$ is $P(Y) = 0.30$ . The analyst confirms that Flaw $X$ and Flaw $Y$ are mutually exclusive events.

Which of the following statements MUST be true regarding these two flaw types? Select all such statements.

The probability that a randomly selected panel has at least one of the two flaws is $0.50$ .Cevap
The conditional probability of a panel having Flaw $X$ given that it has Flaw $Y$ , $P(X \mid Y)$ , is equal to $0$ .Cevap
C
The probability that a randomly selected panel has both Flaw $X$ and Flaw $Y$ is $0.06$ .
D
Flaw $X$ and Flaw $Y$ are independent events.
E
The probability that a randomly selected panel has neither Flaw $X$ nor Flaw $Y$ is $0.94$ .

Cevap

The statement that the probability of having at least one flaw is 0.50 and the statement that the conditional probability P(X | Y) is 0 are both true.

Because Flaw

X

and Flaw

Y

are mutually exclusive, their intersection

P(X \cap Y) = 0

. By the addition rule, the probability of at least one flaw is

P(X \cup Y) = P(X) + P(Y) = 0.20 + 0.30 = 0.50

. Furthermore, the conditional probability

P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0}{0.30} = 0

. Thus, both the statement claiming the union probability is

0.50

and the statement claiming the conditional probability is

0

are correct.

Adım Adım Çözüm

Analyze the definition of mutually exclusive events

Since Flaw

X

and Flaw

Y

are mutually exclusive, they cannot occur simultaneously on the same panel. Therefore,

P(X \cap Y) = 0

By definition, mutually exclusive events have an intersection probability of zero.

Calculate the union probability P(X or Y)

Using the addition rule for mutually exclusive events:

P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = 0.20 + 0.30 - 0 = 0.50

The probability of at least one event occurring is the sum of their individual probabilities when the intersection is zero.

Calculate the conditional probability P(X | Y)

P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0}{0.30} = 0

If Flaw

Y

is known to occur, Flaw

X

cannot occur due to mutual exclusivity.

Evaluate independence between the two events

For independence,

P(X \cap Y)

must equal

P(X) \times P(Y) = 0.20 \times 0.30 = 0.06

. Since

0 \neq 0.06

, the events are dependent.

Two events with non-zero probabilities that are mutually exclusive are always dependent because the occurrence of one guarantees the non-occurrence of the other.

Anahtar Kavram

Mutually Exclusive vs. Independent Events

Soru

Cevap

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Anahtar Kavram

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