There are the combinations that result in a total less than 7 and at least one die showing a 3:
[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36
There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:
1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36
Therefore, the required probability is: 5/36
Answer:
x is 1 y is 1 and a is 1 hope this helped
Answer:
0
Step-by-step explanation:
3*(m-4)-n*(m-4) m = 5, n = 3
3*(5-4)-3*(5-4)
3*(1)-3*(1)
3-3
0
48 divided by sqrt of 9.5 =
48 divided by 3.082207 = 15.57325643605378 is your answer
Step-by-step explanation:
Range of the following:
-3,-2,1
