George and Hamza both roll two standard fair -sided dice. What is the probability that Hamza rolls the same two numbers as George, though not necessarily in the same order?
+4 Hello,
The probability that Hamza rolls the same two numbers as George, though not necessarily in the same order, can be calculated using the concept of combinations.
There are 6 sides on a standard fair-sided die, so the total number of possible outcomes for George's roll is 6 * 6 = 36.
Now, let's consider the number of favorable outcomes where Hamza rolls the same two numbers as George, not necessarily in the same order. In this case, Hamza's roll can be thought of as choosing 2 numbers out of the 6 possibilities, without regard to order. TKMaxxCare
This can be calculated using combinations: C(n, r) = n! / (r! * (n - r)!), where n is the total number of possibilities (6 in this case) and r is the number of choices (2 in this case).
C(6, 2) = 6! / (2! * (6 - 2)!) = 15
So, there are 15 favorable outcomes where Hamza rolls the same two numbers as George, not necessarily in the same order.
The probability is then the ratio of favorable outcomes to total outcomes:
Probability = Favorable Outcomes / Total Outcomes
Probability = 15 / 36
Probability = 5/12
Therefore, the probability that Hamza rolls the same two numbers as George, though not necessarily in the same order, is 5/12 or approximately 0.4167.
