Answer :
When dealing with probabilities of sequential events, the probability of both events happening is the product of the probabilities of each individual event happening.
Here, we are dealing with a six-sided number cube (a standard die) being rolled twice, and we want to find the probability of rolling a 5 on the first roll and then a 3 on the second roll.
1. Probability of rolling a 5 on the first roll:
Since there are 6 faces on the die, and only one of those faces is a 5, the probability of rolling a 5 is:
[tex]\[ P(\text{rolling a 5}) = \frac{1}{6} \][/tex]
2. Probability of rolling a 3 on the second roll:
Similarly, since there are 6 faces on the die, and only one of those faces is a 3, the probability of rolling a 3 is:
[tex]\[ P(\text{rolling a 3}) = \frac{1}{6} \][/tex]
3. Combined probability of both events happening (rolling a 5 first and then rolling a 3):
Since the rolls are independent events, the combined probability is the product of the individual probabilities:
[tex]\[ P(5 \text{ then } 3) = P(\text{rolling a 5}) \times P(\text{rolling a 3}) = \frac{1}{6} \times \frac{1}{6} \][/tex]
Therefore, the expression that can be used to find [tex]\( P(5 \text{ then } 3) \)[/tex] is:
[tex]\[ \boxed{\frac{1}{6} \cdot \frac{1}{6}} \][/tex]
By multiplying these together, we get:
[tex]\[ \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \][/tex]
Thus, the correct answer is the last choice:
[tex]\[ \boxed{\frac{1}{6} \cdot \frac{1}{6}} \][/tex]
Here, we are dealing with a six-sided number cube (a standard die) being rolled twice, and we want to find the probability of rolling a 5 on the first roll and then a 3 on the second roll.
1. Probability of rolling a 5 on the first roll:
Since there are 6 faces on the die, and only one of those faces is a 5, the probability of rolling a 5 is:
[tex]\[ P(\text{rolling a 5}) = \frac{1}{6} \][/tex]
2. Probability of rolling a 3 on the second roll:
Similarly, since there are 6 faces on the die, and only one of those faces is a 3, the probability of rolling a 3 is:
[tex]\[ P(\text{rolling a 3}) = \frac{1}{6} \][/tex]
3. Combined probability of both events happening (rolling a 5 first and then rolling a 3):
Since the rolls are independent events, the combined probability is the product of the individual probabilities:
[tex]\[ P(5 \text{ then } 3) = P(\text{rolling a 5}) \times P(\text{rolling a 3}) = \frac{1}{6} \times \frac{1}{6} \][/tex]
Therefore, the expression that can be used to find [tex]\( P(5 \text{ then } 3) \)[/tex] is:
[tex]\[ \boxed{\frac{1}{6} \cdot \frac{1}{6}} \][/tex]
By multiplying these together, we get:
[tex]\[ \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \][/tex]
Thus, the correct answer is the last choice:
[tex]\[ \boxed{\frac{1}{6} \cdot \frac{1}{6}} \][/tex]
