    # What is the probability of getting consecutive two even numbers on rolling out a dice twice?

When rolling two dice, distinguish between them in some way: a first one and second one, a left and a right, a red and a green, etc. Let (a,b) denote a possible outcome of rolling the two die, with a the number on the top of the first die and b the number on the top of the second die. Note that each of a and b can be any of the integers from 1 through 6. Here is a listing of all the joint possibilities for (a,b):

 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36. The set of all possible outcomes for (a,b) is called the sample space of this probability experiment.

With the sample space now identified, formal probability theory requires that we identify the possible events. These are always subsets of the sample space, and must form a sigma-algebra. In an example such as this, where the sample space is finite because it has only 36 different outcomes, it is perhaps easiest to simply declare ALL subsets of the sample space to be possible events. That will be a sigma-algebra and avoids what might otherwise be an annoying technical difficulty. We make that declaration with this example of two dice.

With the above declaration, the outcomes where the sum of the two dice is equal to 5 form an event. If we call this event E, we have

E={(1,4),(2,3),(3,2),(4,1)}.
Note that we have listed all the ways a first die and second die add up to 5 when we look at their top faces.

Consider next the probability of E, P(E). Here we need more information. If the two dice are fair and independent , each possibility (a,b) is equally likely. Because there are 36 possibilities in all, and the sum of their probabilities must equal 1, each singleton event {(a,b)} is assigned probability equal to 1/36. Because E is composed of 4 such distinct singleton events, P(E)=4/36= 1/9.

In general, when the two dice are fair and independent, the probability of any event is the number of elements in the event divided by 36.

What if the dice aren't fair, or aren't independent of each other? Then each outcome {(a,b)} is assigned a probability (a number in [0,1]) whose sum over all 36 outcomes is equal to 1. These probabilities aren't all equal, and must be estimated by experiment or inferred from other hypotheses about how the dice are related and and how likely each number is on each of the dice. Then the probability of an event such as E is the sum of the probabilities of the singleton events {(a,b)} that make up E.

## Example 1

An unbiased coin is tossed twice.

(a)

List all the possible outcomes. The possible outcomes are:

So there are 4 possible outcomes that are all equally likely to occur as the coin is not biased.

(b)

What is the probability of obtaining two heads? There is only one way of obtaining 2 heads, so:

(c)

What is the probability of obtaining a head and a tail in any order? There are two ways of obtaining a head and a tail, H T and T H, so:

 p(a head and a tail) = =

## Example 2

A red dice and a blue dice, both unbiased, are rolled at the same time. The scores on the two dice are then added together.

(a)

Use a table to show all the possible outcomes. The following table shows all of the 36 possible outcomes:

Red Dice

Blue Dice
1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12

(b)

What is the probability of obtaining:

(i) a score of 5, There are 4 ways of scoring 5, so:

 p(5) = =

(ii) a score which is greater than 3, There are 33 ways of obtaining a score greater than 3, so:

 p(greater than 3) = =

(iii) a score which is an even number? There are 18 ways of obtaining a score which is an even number, so:

 p(even score) = =

## Example 3

A card is taken at random from a pack of 52 playing cards, and then replaced. A second card is then drawn at random from the pack.
Use a tree diagram to determine the probability that: We first note that, for a single card drawn from the pack,

 p(Diamond) = = and p(not Diamond) = = .

We put these probabilities on the branches of the tree diagram below: Note also that the probability for each combination, for example, two Diamonds, is determined by multiplying the probabilities along the branches.

(a)

both cards are Diamonds, (b)

at least one card is a Diamond, p(at least one Diamond) = + + =

(c)

exactly one card is a Diamond, p(exactly one Diamond) = + = =

(d)

neither card is a Diamond. p(neither card a Diamond) =

## Exercises

Question 1

The faces of an unbiased dice are painted so that 2 are red, 2 are blue and 2 are yellow. The dice is rolled twice. Three of the possible outcomes are listed below:

(a)

List all 9 possible outcomes.

(b)

What is the probability that:

(i) both faces are red, both faces are the same colour, the faces are of different colours?

Question 2

A spinner is marked with the letters A, B, C and D, so that each letter is equally likely to be obtained. The spinner is spun twice.

(a)

List the 16 possible outcomes.

(b)

What is the probability that:

(i) A is obtained twice, A is obtained at least once, both letters are the same, the letter B is not obtained at all?

Question 6
 A coin is biased so that the probability of obtaining a head is
 and the probability of obtaining a tail is .

(a)

Complete the following tree diagram to show the possible outcomes and probabilities if the coin is tossed twice. (b)

What is the probability of obtaining:

(i) 2 heads, at least one head, 2 tails, exactly 1 tail ?

Question 8

A card is taken at random from a pack of 52 playing cards. It is replaced and a second card is then taken at random from the pack. A card is said to be a 'Royal' card if it is a King, Queen or Jack.
Use a tree diagram to calculate the probability that: (a) both cards are Royals, one card is a Royal, at least one card is a Royal, neither card is a Royal.

Question 9
 The probability that a school bus is late on any day is .

Use a tree diagram to calculate the probability that on two consecutive days, the bus is: (a) late twice, late once, never late.

Question 10
 The probability that a piece of bread burns in a toaster is .
Two slices of bread are toasted, one after the other.

(a) Use a tree diagram to calculate the probability that at least one of these slices of bread burns in the toaster.

 p(at least one slice burnt) =

(b) Extend your tree diagram to include toasting 3 slices, one at a time. Calculate the probability of at least one slice burning in the toaster.

 p(at least one slice burnt) =

Question 12

I have two fair dice. Each of the dice is numbered 1 to 6.

(a)

The probability that I will throw double 6 (both dice showing number 6) is

What is the probability that I will not throw double 6 ?

(b)

I start again and throw both dice.

(c)

What is the probability that I will throw double 3 (both dice showing number 3) ?

(d)

What is the probability that I will throw a double? (It could be double 1 or double 2 or any other double.)

Question 13

On a road there are two sets of traffic lights. The traffic lights work independently.
For each set of traffic lights, the probability that a driver will have to stop is 0.7.

(a)

A woman is going to drive along the road.

(i)

What is the probability that she will have to stop at both sets of traffic lights?

0.7 × 0.7 = 0.49

(ii)

What is the probability that she will have to stop at only one of the two sets of traffic lights?

(0.7 × 0.3) + (0.3 × 0.7) = 0.42

(b)

In one year, a man drives 200 times along the road. Calculate an estimate of the number of times he drives through both sets of traffic lights without stopping.

p(drives through both sets of lights without stopping) = 0.3 × 0.3 = 0.09,
so the estimated number of times he goes through unstopped = 200 × 0.09 = 18.

Question 14

100 students were asked whether they studied French or German.

 Results: 27 students studied both French and German.

(a)

What is the probability that a student chosen at random will study only one of the languages?

Note: Write the solution as a decimal or percentage

(b)

What is the probability that a student who is studying German is also studying French?

(c)

Two of the 100 students are chosen at random.
From the following calculations, choose one which shows the probability that both students study French and German.

Note: Choose a calculation by clicking on it.

Question 15

A company makes computer disks. It tested a random sample of the disks from a large batch. The company calculated the probability of any disk being defective as 0.025.
Glenda buys 2 disks.

(a)

Calculate the probability that both disks are defective.

(b)

Calculate the probability that only one of the disks is defective.

(c)

The company found 3 defective disks in the sample they tested.
How many disks were likely to have been tested?

Question 16

 On a tropical island the probability of it raining on the first day of the rainy season is .
 If it does not rain on the first day, the probability of it raining on the second day is .
 If it rains on the first day, the probability of it raining more than 10 mm on the first day is .
 If it rains on the second day but not on the first day, the probability of it raining more than 10 mm is .

You may find it helpful to copy and complete the tree diagram before answering the questions. (a)

What is the probability that it rains more than 10 mm on the second day, and does not rain on the first?

(b)

What is the probability that it has rained by the end of the second day of the rainy season?

(c)

Is it possible to work out the probability of rain on both days from the information given?

Because we are not given the probability that it rains in the second day if it rains on the first.

Question 17

Pupils at a school invented a word game called Wordo. They tried it out with a large sample of people and found that the probability of winning Wordo was 0.6.
The pupils invented another word game, Lango. The same sample who had played Wordo then played Lango. The pupils drew this tree diagram to show the probabilities of winning. (a)

What was the probability of someone from the sample winning Lango?

(0.6 × 0.8) + (0.4 × 0.55) = 0.7

(b)

What was the probability of someone from the sample winning only one of the two word games?

(0.6 × 0.2) + (0.4 × 0.55) = 0.34

(c)

The pupils also invented a dice game. They tried it out with the same sample of people who had already played Wordo and Lango.
The probability of winning the dice game was 0.9. This was found to be independent of the probabilities for Wordo and Lango.
Calculate the probability of someone from the sample winning two out of these three games.

(0.6 × 0.8 × 0.1) + (0.6 × 0.2 × 0.9) + (0.4 × 0.55 × 0.9) = 0.354

(d)

Calculate the probability of someone from the sample winning only one of these three games.

(0.6 × 0.2 × 0.1) + (0.4 × 0.55 × 0.1) + (0.4 × 0.45 × 0.9) = 0.196 