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Two six-sided dice are tossed.

Event A: The first die does NOT land on 5.

Event B: The second die lands on 4.

What is the probability that both events will occur?

For independent events: [tex]P(A \text{ and } B) = P(A) \cdot P(B)[/tex]

[tex]P(A \text{ and } B) = \square[/tex]

Give your answer in simplest form.


Sagot :

To find the probability that both events A and B will occur, we'll follow these steps:

1. Determine the probability of Event A: The first die does NOT land on 5.

Since a six-sided die has 6 outcomes (1, 2, 3, 4, 5, 6) and only one of these outcomes is a 5, there are 5 favorable outcomes where the first die does not land on 5 (i.e., 1, 2, 3, 4, 6).

Hence, the probability of Event A occurring is:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{6} \][/tex]

2. Determine the probability of Event B: The second die lands on 4.

Similarly, since a six-sided die has 6 outcomes, and only one of these outcomes is a 4, there is exactly 1 favorable outcome.

Thus, the probability of Event B occurring is:
[tex]\[ P(B) = \frac{1}{6} \][/tex]

3. Find the probability that both events A and B will occur.

Since the events are independent, the probability that both events occur is the product of the probabilities of each event:

[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) \][/tex]

Substituting the values of [tex]\( P(A) \)[/tex] and [tex]\( P(B) \)[/tex]:

[tex]\[ P(A \text{ and } B) = \frac{5}{6} \cdot \frac{1}{6} \][/tex]

Simplify the expression:

[tex]\[ P(A \text{ and } B) = \frac{5 \times 1}{6 \times 6} = \frac{5}{36} \][/tex]

Therefore, the probability that both events A and B will occur is:
[tex]\[ P(A \text{ and } B) = \frac{5}{36} \][/tex]