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
when the two books were published
Answer:
=40b+24
Step-by-step explanation:
Answer:
£2.10
Step-by-step explanation:
1.80 divided by .12 = .15
.15 times 7 = 1.05
1.05 divided by 3 = .35
.35 times 5 = 2.10
-5x - 49 > 113
Add 49 to both sides
-5x > 162
Divide both sides by -5
x > -32.4
Hope this helps!
