A die is thrown twice. Find the probability that the number obtained in the first throw is smaller than that in the second throw.
just add them up, considering the 1st throw:
6: 0
5: 1/6
4: 2/6
...
1: 5/6
sum: (1/6)(15/6) = 5/12
since the chance of throwing each of the 1st numbers is 1/6.
To find the probability that the number obtained in the first throw is smaller than that in the second throw, we can list all possible outcomes and count the favorable outcomes.
Step 1: Determine the sample space (all possible outcomes)
When a fair die is thrown twice, the possible outcomes for each throw are numbers 1, 2, 3, 4, 5, or 6. Since there are six possible outcomes for each throw, the sample space for this experiment is 6 x 6 = 36. So there are 36 possible outcomes.
Step 2: Identify the favorable outcomes
Out of the 36 possible outcomes, there are 6 cases where the number obtained in the first throw is smaller than that in the second throw: (1,2), (1,3), (1,4), (1,5), (1,6), (2,6). So, there are 6 favorable outcomes.
Step 3: Calculate the probability
The probability of an event is given by the ratio of favorable outcomes to the total number of outcomes. In this case, the probability is:
Probability = Favorable Outcomes / Total Outcomes
Probability = 6 / 36
Simplifying the fraction, we find:
Probability = 1 / 6
Therefore, the probability that the number obtained in the first throw is smaller than that in the second throw is 1/6 or approximately 0.1667.