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Explain why the statement is incorrect. A red die and a green die are tossed. The probability that a 6 will appear uppermost on the red die is \(\frac{1}{6}\), and the probability that a 1 will appear uppermost on the green die is \(\frac{1}{6}\). Hence, the probability that the red die will show a 6 or the green die will show a 1 is \(\frac{1}{6}+\frac{1}{6}\).

Short Answer

Expert verified
The original statement is actually correct because the events of the red die showing a 6 and the green die showing a 1 are independent. Therefore, the probability of either event occurring is the sum of their individual probabilities: P(A ∪ B) = P(A) + P(B) = \(\frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}\).

Step by step solution

01

Define the events and probabilities

Let A be the event that the red die shows a 6, and B be the event that the green die shows a 1. The probability of A (P(A)) is given as \(\frac{1}{6}\), and the probability of B (P(B)) is given as \(\frac{1}{6}\).
02

Determine if the events are independent

Two events are independent if the occurrence of one event does not affect the occurrence of the other event. Since the outcomes of rolling the red and green dice are not related in any way, we can conclude that the events A and B are independent.
03

Calculate the probability of either event occurring

Since the events are indeed independent, the probability of either A or B occurring (A 6 on the red die or a 1 on the green die) is the sum of their individual probabilities. Therefore, the probability of either event occurring is: P(A ∪ B) = P(A) + P(B) = \(\frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}\). The original statement is thus correct. The probability of the red die showing a 6 or the green die showing a 1 is indeed \(\frac{1}{6}+\frac{1}{6}\).

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