If you randomly choose two different numbers from 1 to 10, what is the probability that one number is one more than the other?
If they are different numbers, the probability that one is greater than the other one is equal to 1.
Goodness.
you could have picked 1-2, 2-3, 3-4, 4-5, 5-6 , 6-7, 7-8, 8-9, 9-10 or their reverses.
so prob(your event) = 18/100 = 9/50
1/5
tangina basta yan ung tama
There are 10!/8!2!=45 possible combinations that you will get given that you draw two numbers at a time. Then, there are possible combinations you could draw where numbers are consecutive e.g. (1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,10) and vice-versa for a total of 9 possible combinations. Thus, the probability that one number is one more than the other is 9/45 or 1/5. Hope it helps. :)
To find the probability that one number is one more than the other when choosing two different numbers from 1 to 10, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, let's find the total number of possible outcomes. When choosing two different numbers, the first number can be any of the 10 options (1 to 10), and the second number can be any of the remaining 9 options (as it must be different from the first number). Therefore, the total number of possible outcomes is 10 * 9 = 90.
Now, let's determine the number of favorable outcomes, i.e., the number of outcomes where one number is one more than the other. There are two cases to consider:
1. The first number is one more than the second number.
In this case, we have the following pairs: (2, 1), (3, 2), (4, 3), ..., (10, 9). There are 9 pairs in total.
2. The second number is one more than the first number.
In this case, we have the following pairs: (1, 2), (2, 3), (3, 4), ..., (9, 10). Again, there are 9 pairs.
Therefore, the total number of favorable outcomes is 9 + 9 = 18.
Finally, to calculate the probability, we divide the number of favorable outcomes by the total number of possible outcomes:
Probability = Favorable outcomes / Total outcomes
Probability = 18 / 90 = 0.2, or 20%.
So, the probability that one number is one more than the other is 20% when randomly choosing two different numbers from 1 to 10.