A white die and a black die is thrown. What is the probability that the number on the white die exceeds the number on the black die?
1/2
cases we care about: White is 1st number , Black is 2nd
21
31 32
41 42 43
51 52 53 54
61 62 63 64 65
prob of your event = 15/36 = 5/12
To find the probability that the number on the white die exceeds the number on the black die, we need to consider all possible outcomes of throwing the two dice.
Step 1: Determine the total number of outcomes when throwing two dice.
Each die has six sides, so the total number of possible outcomes for throwing one die is 6. Since we have two dice, the total number of outcomes when throwing both dice is 6 * 6 = 36.
Step 2: Identify the favorable outcomes.
In this case, the favorable outcomes are when the number on the white die exceeds the number on the black die. We'll list and count these outcomes:
White die: 2, black die: 1
White die: 3, black die: 1
White die: 3, black die: 2
White die: 4, black die: 1
White die: 4, black die: 2
White die: 4, black die: 3
White die: 5, black die: 1
White die: 5, black die: 2
White die: 5, black die: 3
White die: 5, black die: 4
White die: 6, black die: 1
White die: 6, black die: 2
White die: 6, black die: 3
White die: 6, black die: 4
White die: 6, black die: 5
There are 15 favorable outcomes in total.
Step 3: Calculate the probability.
The probability is the ratio of favorable outcomes to the total number of outcomes.
Probability = Number of favorable outcomes / Total number of outcomes
In this case, the probability is 15 / 36.
Therefore, the probability that the number on the white die exceeds the number on the black die is 15/36, which simplifies to 5/12 or approximately 0.4167.