What is the probability of rolling two number cubes and having the first cube land on 2 and the second cube land on 3?
I got 2/6 * 3/6 = 6/36 which becomes 1/6
1/6 probability is my answer correct?
No.
The probability of the first die rolling a 2 is 1/6.
The probability of the second die rolling a 3 is 1/6.
The combined probability is 1/36.
There is only one #2 and one #3 on each die.
1/6 * 1/6 = 1/36
Yes, your answer is correct. To find the probability of two independent events occurring, you multiply the probabilities of each event occurring. For rolling a number cube, there are six equally likely outcomes (numbers 1 to 6). So the probability of rolling a 2 on the first cube is 1/6, and the probability of rolling a 3 on the second cube is also 1/6. Therefore, the probability of both events occurring is (1/6) * (1/6) = 1/36.
Yes, your answer is correct. The probability of rolling a specific number on a fair six-sided number cube is 1/6.
To calculate the probability of both events occurring, you need to multiply the probabilities of each event happening individually. In this case, the probability of the first cube landing on 2 is 1/6 (as there is only one favorable outcome out of six possible outcomes), and the probability of the second cube landing on 3 is also 1/6. Multiplying these probabilities together gives you (1/6) * (1/6) = 1/36.
Since you are asking for the probability of both cubes landing on specific numbers in any order (e.g., first cube on 2 and second cube on 3 or first cube on 3 and second cube on 2), you need to multiply the probability by 2, giving you a final probability of (1/36) * 2 = 1/18.