Four fair sided dice, each with faces marked 1,2,3,4,5,6, are thrown on 7 occasions.Find the probability that the numbers shown on the four dice add up to 5 on exactly one or two of the 7 occasions.

there are 6^4 possible outcomes with 4 dice

there are 4 ways to get a sum of 5
... 3 dice show one, and the 4th shows 2

f = p(5) = 4 / (6^4) ... n = p(not 5) = 1 - [4 / (6^4)]

this is binomial , 5 or (not 5) over 7 throws

(n + f)^7 = n^7 + 7 n^6 f + 21 n^5 f^2 + ... + 7 n f^6 + f^7

the answer is the sum of the 2nd and 3rd terms

To find the probability that the numbers shown on the four dice add up to 5 on exactly one or two of the 7 occasions, we need to count the number of favorable outcomes and divide it by the total number of possible outcomes.

1. Let's consider the case when the numbers add up to 5 on exactly one occasion:
- We need one occasion where the numbers add up to 5.
- To achieve this, we can have either of the following combinations for the four dice: (1,1,1,2), (1,1,2,1), (1,2,1,1), or (2,1,1,1).
- Since there are 4 ways to arrange the dice rolls, the number of favorable outcomes in this case is 4.

2. Now let's consider the case when the numbers add up to 5 on exactly two occasions:
- We need two occasions where the numbers add up to 5.
- To achieve this, we can have the following combinations for the four dice: (1,1,1,2), (1,1,2,1), (1,2,1,1), or (2,1,1,1).
- There are 4 ways to arrange the dice rolls as the first occasion and 4 ways for the second occasion, giving us a total of 4 * 4 = 16 favorable outcomes in this case.

3. To find the total number of possible outcomes, we need to calculate the number of ways we can throw four fair dice on 7 occasions. Since each die has 6 faces and we have four dice, the total number of possible outcomes is 6^4 * 7.

4. The final step is to calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
- Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
- Probability = (4 + 16) / (6^4 * 7)

Simplifying this expression will give us the final probability.

To find the probability that the numbers shown on the dice add up to 5 on exactly one or two of the 7 occasions, we need to consider the different ways this can occur.

Let's break it down step by step:

Step 1: Find the total number of possible outcomes when throwing four fair-sided dice. Since each die has 6 possible outcomes, the total number of possible outcomes when throwing four dice is 6^4 = 1,296.

Step 2: Determine the total number of favorable outcomes where the numbers on the dice add up to 5 on exactly one or two occasions. To do this, we'll count the different cases separately.

Case 1: Exactly one occasion has the sum of 5.
There are three possible ways this can happen:
a) The first die shows a 1 and the others show any numbers other than 4.
b) The second die shows a 1 and the others show any numbers other than 4.
c) The third die shows a 1 and the others show any numbers other than 4.
Each of these cases gives us 3 favorable outcomes.

Case 2: Exactly two occasions have the sum of 5.
There are two possible ways this can happen:
a) The first and second dice show a sum of 5, while the remaining two dice show other numbers.
b) The first and third dice show a sum of 5, while the remaining two dice show other numbers.
Each of these cases gives us 3 favorable outcomes.

So, the total number of favorable outcomes in the given cases is 3 + 3 + 3 + 3 = 12.

Step 3: Calculate the probability by dividing the total number of favorable outcomes (12) by the total number of possible outcomes (1,296).

Probability = Favorable outcomes / Total outcomes
Probability = 12 / 1,296

Simplifying the fraction, we get:
Probability = 1 / 108

Therefore, the probability that the numbers shown on the four dice add up to 5 on exactly one or two of the 7 occasions is 1/108.