Question

1. A round in the game Yahtzee begins by rolling five fair dice. Find the probability of rolling a:

a. one pair (ex 33421 but not 33441), two pair (ex 33441 but not 33444), and three of a kind (ex 24252 but not 24242)

2. Consider a 10x10 matrix that consists of all zeros. ten elements of the matrix are selected at random and their value is changed from a zero to a one. find the probability that the ones fall in a line (row-wise, column-wise, or diagonally)

3. Five fair dice are rolled simultaneously. Find the probability that the total number of spots on the up faces totals 28 or more

Answer 1

Total outcome= (each dice can have 6 outcomes)

Now there can be any of the 6 pairs: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) and other 3 distinct numbers

We will consider the positions of identical numbers. Since we are rolling 5 dice there are ways to place two identical faced dice. Remaining 3 dices will have other different numbers.

Among these 3 dices there are 5 outcomes for the first of the remaining dice, 4 for the second and 3 for the last one

All total we have ways of getting a particular pair. and we have 6 pairs.

Hence the total number of getting exactly one pair is:

Total outcome= (each dice can have 6 outcomes)

There are six choose two ways to get two numbers to be paired: =15
There are four numbers to choose from for the singleton: 4

Now there is total of 60 combinations.

Any of the 5 dice can take that number then the number of ways becomes (5*60)

And that particular dice can be placed in 6 different ways in between the other dices

So we get 60*5*6=1800

First pick the number for 3 of a kind That can be in 6 ways
Now choose two singletons: =10
So again 60 combinations.

Remaining 2 dices will have other different numbers.

Among these 2 dices there are 5 outcomes for the first 4 for the last one.

Hence the total number of getting 3 of a kind is (60*5*4)=1200

In a 10X10 matrix there are 100 cells. According to the problem 100 "0"s

There are 10 rows 10 columns and 2 diagonals. For all 1s to fall in a line we must choose either one of the rows or one of the columns or one of the diagonals. Then there is all total (10+10+2)=22 ways that all the 1s will fall in a line.

And total way of choosing 10 cell is ways

the probability that the ones fall in a line is

  

If five fair dice are rolled simultaneously maximum we get 5 6s which adds up to 30.

Therefore we have to find P(sum of the faces adds up to 28, 29 and 30)

=P(sum is 28)+P(sum is 29)+P(sum is 30)

We can get 30 only in one way

Now to get 29 one of the dice must face 5 and others 6. Any of the 6 dice can have 5

Then total number of getting 29 is 6

To get 28 we have 2 cases

1. 2 dice face 5 and others 6: This can have =15 ways.

2. one dice face 4 and others 6: Any of the 6 dice can have 4.Then total ways are 6

Then total number of getting 28 is (15+6)=21.

P(sum is 28)=   P(sum is 29)= P(sum is 30)=

the probability that the total number of spots on the up faces totals 28 or more is: