probability problems hope someone is amazing at these?

With any probability question, you need to consider the total number of possible outcomes. You then work out how many of those fit the "rule" you are finding the probability of.

If you roll the die the first time, there are 6 possible results (1-6).

For each of those 6 possible outcomes, there are also 6 possible outcomes for the second roll. That means there are 6 * 6 = 36 possible outcomes of rolling a die twice.

Now let's see how many of them work for this rule. The sum of 2 rolls of a die, can be any number from 2 (both ones), to 12 (both sixes).

If you get a 5, then you could have rolled:

1 then 4

2 then 3

3 then 2

4 then 1.

There are 4 possible outcomes of the 2 rolls that fit your "rule".

So, the probability is 4/36 = 1/9

Starting with the definition of probability of an event Pr[x]

= Number of outcomes favorable / Total number of possible outcomes.

The denominator is easy. Roll a dice twice, total number of possible outcomes = 6*6 = 36

Now, how many ways can the sum of two consecutive die throw equal 5?

Just enumerate them: { 1,4 2,3 3,2 4,1 }

prob = 4/36

Just by taking this problem brute force,

Each roll of the die is independent and random, so

we have an option of rolling a (1,1) (1,2) (1,3) (1,4)... etc.

Since the first roll is different from the second, both (3,4) and (4,3) count

To get a 5, we need a 1 and 4, or a 2 and 3

This is represented by (1,4), (4,1), (2,3), (3,2) or 4 combinations.

There are a total of 5*5 combinations, or 25

Since 4 of them work, the probability of getting a sum of 5 is

4/25, or 16%

P(huge type is divisble through 2)= (6P5 x 4)/7P6= 4/7 that is becuase there are 5 numbers which will be chosen from the 6. and the ultimate digit must be then back it cam be chosen from the 4 achieveable even numbers. P(23)= 5P4x5/ 7P6= 5/40 2 that is because there are 5P4 wasy to go back to a decision the different 4 numbers to boot 23 and its cases 5 because theres 5 places the position u would position the 23. wish this helps!

1) When you roll a die once, you get {1,2,3,4,5,6}

2) When you roll second time also you get the same

3) But combining both you get

{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1) (2,2), ------------- (6,6)}

Thus you have 36 elements.

4) Of these favorable to get sum 5 = {(1,4), (2,3), (3,2), (4,1)}

Thus it has 4 elements

Hence the probability = 4/36 = 1/9.