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:
w + r 》 4 (greater than/equal to 4)
270w + 650r 》 1500
I'm pretty sure k doesn't have a value it is no solution
Answer:
anything above 7 or not equivalent to 7 can be the answer
Step-by-step explanation:
8÷7=a undifined number but if it is 7÷7 its 1 and if you got 7÷1 it is 7 so anything other than those answers should be it
