Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
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]
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.