When two dice are thrown simultaneously the probability of getting the same even number on both the dice is?

I'm fairly certain that the probability of both dice returning an even number is $1/4$.

I got this by saying that since these are independent events, with each die returning an even number being $1/2$, then the probability of both being even is $1/2 \times 1/2 = 1/4$.

Further, there are 36 outcomes, and all possible even number combinations are $(2, 2), (2, 4), (2, 6), (4, 4), (4, 6), (6, 6), (6, 4), (6, 2), (4, 2)$. There are nine of them and $9/36 = 1/4$

What I can't seem to get over, is that there are an equal number of odd and even numbers, so, why is the answer not $1/2$?

I know that it's not one half, but I can't explain why.

Answer

Hint: Calculate the number of possible outcomes for throwing two dice. Calculate the number of favourable outcomes for each of the cases. Use the fact that the probability of any event is the ratio of the number of favourable outcomes and the number of possible outcomes to calculate the probability of each of the events.

Complete step-by-step solution -

Probability is a measure of the possibility of how likely an event will occur. It is a value between 0 and 1 which shows us how favorable is the occurrence of a condition. If the probability of an event is nearer to 0, let’s say 0.2 or 0.13 then the possibility of its occurrence is less. Whereas if the probability of an event is nearer to 1, lets say 0.92 or 0.88 then it is much favourable to occur.

Probability of an event

The probability of an event can be defined as a number of favorable outcomes upon the total number of outcomes.

P(A) = Number of favorable outcomes / Total number of outcomes

Some terms related to probability

• Experiment: An experiment is any action or set of action performed to determine the probability of an event. The result of action performed is random or uncertain. e.g. Tossing a coin, rolling dice, etc.
• Event: An event can be defined as certain condition which can happen while performing an experiment. e.g. getting head while tossing a coin, getting even number while rolling dice, etc.
• Sample Space: It is set of all the possible outcomes which happens after performing an experiment. e.g. Sample Space of tossing a coin = {H,T} and Sample Space of rolling a dice = {1,2,3,4,5,6}, so on.
• Sample Point: It is a part of sample space which contains one of the outcomes from Sample Space. e.g. Getting 1 while rolling dice, getting an ace of Spades while drawing a card from pack of cards, etc.
• Types of Events: There are majorly four kinds of events that are-
  • Complimentary events- It is used to find probability of not happening of an event. It is denoted by ( ‘ ) symbol. If event is denoted by A, then complimentary of event is A’. e.g. probability of not getting 2 while rolling a dice. It can be calculated by subtracting normal probability from 1 i.e. P(A’) = 1 – P(A)
  • Impossible event- Impossible event is a type of event which can never happen. The probability of Impossible event is 0. e.g. getting a number 8 while rolling a dice.
  • Certain event- Certain event is a type of event which always happen. The probability of a certain event is 1. e.g. getting a head or a tail after tossing a coin.
  • Equally Likely events- Events whose probability of occurrence are equal i.e. they are equally likely to happen. The value of probability of such events are same. e.g. getting a head and getting a tail both have 50% probability.

When two dice are rolled what is the probability of getting same number on both?

Since, the number of outcomes while rolling a dice = 6

Number of outcomes while rolling two dice = 62

= 36

The Sample Space for rolling a die is given as,

{(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)}

Sample points of getting same number on both dice- (1,1) ,(2,2) ,(3,3) ,(4,4) ,(5,5) & (6,6).

Thus, the number of favourable outcomes = 6

Total number of outcomes = 36

P (getting same number on both dice) = 6/36

= 1/6

Hence, the probability of getting same number on both the dice is 1/6.

Sample Questions

Question 1: Find the probability of getting odd number on first dice and even number on other dice when two dice are thrown simultaneously.

Answer: 

Total number of outcomes = 36

Sample Space :

{(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)}

In 9 outcomes we will get odd number on first dice and even number on second dice.

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

Question 2: If two dice are thrown together then find the probability of getting 1 or 2 on either of the dice.

Answer: 

Total number of outcomes = 36

Sample Space :

{(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)}

From above sample space it is clear that there are total 20 possibilities in which 1 or 2 appears on either of the dice.

So, required possibility = 20/36 = 5/9

Question 3: In an event 2 dice are thrown simultaneously. Find the probability of getting prime number on first dice.

Answer:

Total number of possibilities = 36

Sample Space :

{(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)}

Since, 2, 3 and 5 are prime number which appear on first dice in 2nd, 3rd and 5th row of sample space respectively.

So, number favourable sample points = 18

Required probability = 18/36

Question 4: Three coins are tossed together find the probability of getting at least one head and one tail.

Answer: 

Number of possibilities while tossing a coin = 2

Number of possibilities while tossing 3 coins together = 23

                                                                                    = 8

Sample Space : 

{ (H,H,H) , (H,H,T) , (H,T,H) , (H,T,T) , 

(T,H,H) , (T,H,T) , (T,T,H) , (T,T,T) }

6 sample points are having head and tail both.

P(E) = 6/8 

= 3/4

Question 5: Find the probability of getting at least two tails when a coin is tossed three times.

Answer: 

Total number of outcomes = 8

Sample Space :

{(H,H,H) , (H,H,T) , (H,T,H) , (H,T,T) ,

(T,H,H) , (T,H,T) , (T,T,H) , (T,T,T)}

Number of favourable outcomes = 4

Probability = 4/8

= 1/2