What is the probability of getting 2 atleast in one uppermost face when two dice are thrown?

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

Probability for rolling two dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each die.

What is the probability of getting 2 atleast in one uppermost face when two dice are thrown?

When two dice are thrown simultaneously, thus number of event can be 62 = 36 because each die has 1 to 6 number on its faces. Then the possible outcomes are shown in the below table.

Probability – Sample space for two dice (outcomes):

What is the probability of getting 2 atleast in one uppermost face when two dice are thrown?

Note: 

(i) The outcomes (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) are called doublets.

(ii) The pair (1, 2) and (2, 1) are different outcomes.

Worked-out problems involving probability for rolling two dice:

1. Two dice are rolled. Let A, B, C be the events of getting a sum of 2, a sum of 3 and a sum of 4 respectively. Then, show that

(i) A is a simple event

(ii) B and C are compound events

(iii) A and B are mutually exclusive

Solution:

Clearly, we haveA = {(1, 1)}, B = {(1, 2), (2, 1)} and C = {(1, 3), (3, 1), (2, 2)}.

(i) Since A consists of a single sample point, it is a simple event.

(ii) Since both B and C contain more than one sample point, each one of them is a compound event.

(iii) Since A ∩ B = ∅, A and B are mutually exclusive.

2. Two dice are rolled. A is the event that the sum of the numbers shown on the two dice is 5, and B is the event that at least one of the dice shows up a 3. Are the two events (i) mutually exclusive, (ii) exhaustive? Give arguments in support of your answer.

Solution:

When two dice are rolled, we have n(S) = (6 × 6) = 36.

Now, A = {(1, 4), (2, 3), (4, 1), (3, 2)}, and

B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1,3), (2, 3), (4, 3), (5, 3), (6, 3)}

(i) A ∩ B = {(2, 3), (3, 2)} ≠ ∅.

Hence, A and B are not mutually exclusive.

(ii) Also, A ∪ B ≠ S.

Therefore, A and B are not exhaustive events.

More examples related to the questions on the probabilities for throwing two dice.

3. Two dice are thrown simultaneously. Find the probability of:

(i) getting six as a product

(ii) getting sum ≤ 3

(iii) getting sum ≤ 10

(iv) getting a doublet

(v) getting a sum of 8

(vi) getting sum divisible by 5

(vii) getting sum of atleast 11

(viii) getting a multiple of 3 as the sum

(ix) getting a total of atleast 10

(x) getting an even number as the sum

(xi) getting a prime number as the sum

(xii) getting a doublet of even numbers

(xiii) getting a multiple of 2 on one die and a multiple of 3 on the other die

Solution: 

Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces. We know that in a single thrown of two different dice, the total number of possible outcomes is (6 × 6) = 36.

(i) getting six as a product:

Let E1 = event of getting six as a product. The number whose product is six will be E1 = [(1, 6), (2, 3), (3, 2), (6, 1)] = 4

Therefore, probability of getting ‘six as a product’

               Number of favorable outcomes
P(E1) =     Total number of possible outcome       = 4/36       = 1/9

(ii) getting sum ≤ 3:

Let E2 = event of getting sum ≤ 3. The number whose sum ≤ 3 will be E2 = [(1, 1), (1, 2), (2, 1)] = 3

Therefore, probability of getting ‘sum ≤ 3’

               Number of favorable outcomes
P(E2) =     Total number of possible outcome       = 3/36       = 1/12

(iii) getting sum ≤ 10:

Let E3 = event of getting sum ≤ 10. The number whose sum ≤ 10 will be E3 =

[(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),

(6, 1), (6, 2), (6, 3), (6, 4)] = 33

Therefore, probability of getting ‘sum ≤ 10’

               Number of favorable outcomes
P(E3) =     Total number of possible outcome       = 33/36       = 11/12

(iv) getting a doublet: Let E4 = event of getting a doublet. The number which doublet will be E4 = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] = 6

Therefore, probability of getting ‘a doublet’

               Number of favorable outcomes
P(E4) =     Total number of possible outcome       = 6/36       = 1/6

(v) getting a sum of 8:

Let E5 = event of getting a sum of 8. The number which is a sum of 8 will be E5 = [(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)] = 5

Therefore, probability of getting ‘a sum of 8’

               Number of favorable outcomes
P(E5) =     Total number of possible outcome       = 5/36

(vi) getting sum divisible by 5:

Let E6 = event of getting sum divisible by 5. The number whose sum divisible by 5 will be E6 = [(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)] = 7

Therefore, probability of getting ‘sum divisible by 5’

               Number of favorable outcomes
P(E6) =     Total number of possible outcome       = 7/36

(vii) getting sum of atleast 11:

Let E7 = event of getting sum of atleast 11. The events of the sum of atleast 11 will be E7 = [(5, 6), (6, 5), (6, 6)] = 3

Therefore, probability of getting ‘sum of atleast 11’

               Number of favorable outcomes
P(E7) =     Total number of possible outcome       = 3/36       = 1/12

(viii) getting a multiple of 3 as the sum:

Let E8 = event of getting a multiple of 3 as the sum. The events of a multiple of 3 as the sum will be E8 = [(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3) (6, 6)] = 12

Therefore, probability of getting ‘a multiple of 3 as the sum’

               Number of favorable outcomes
P(E8) =     Total number of possible outcome       = 12/36       = 1/3

(ix) getting a total of atleast 10:

Let E9 = event of getting a total of atleast 10. The events of a total of atleast 10 will be E9 = [(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)] = 6

Therefore, probability of getting ‘a total of atleast 10’

               Number of favorable outcomes
P(E9) =     Total number of possible outcome       = 6/36       = 1/6

(x) getting an even number as the sum:

Let E10 = event of getting an even number as the sum. The events of an even number as the sum will be E10 = [(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 3), (3, 1), (3, 5), (4, 4), (4, 2), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)] = 18

Therefore, probability of getting ‘an even number as the sum

               Number of favorable outcomes
P(E10) =     Total number of possible outcome       = 18/36       = 1/2

(xi) getting a prime number as the sum:

Let E11 = event of getting a prime number as the sum. The events of a prime number as the sum will be E11 = [(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)] = 15

Therefore, probability of getting ‘a prime number as the sum’

               Number of favorable outcomes
P(E11) =     Total number of possible outcome       = 15/36       = 5/12

(xii) getting a doublet of even numbers:

Let E12 = event of getting a doublet of even numbers. The events of a doublet of even numbers will be E12 = [(2, 2), (4, 4), (6, 6)] = 3

Therefore, probability of getting ‘a doublet of even numbers’

               Number of favorable outcomes
P(E12) =     Total number of possible outcome       = 3/36       = 1/12

(xiii) getting a multiple of 2 on one die and a multiple of 3 on the other die:

Let E13 = event of getting a multiple of 2 on one die and a multiple of 3 on the other die. The events of a multiple of 2 on one die and a multiple of 3 on the other die will be E13 = [(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)] = 11

Therefore, probability of getting ‘a multiple of 2 on one die and a multiple of 3 on the other die’

               Number of favorable outcomes
P(E13) =     Total number of possible outcome       = 11/36

4. Two dice are thrown. Find (i) the odds in favour of getting the sum 5, and (ii) the odds against getting the sum 6.

Solution:

We know that in a single thrown of two die, the total number of possible outcomes is (6 × 6) = 36.

Let S be the sample space. Then, n(S) = 36.

(i) the odds in favour of getting the sum 5:

Let E1 be the event of getting the sum 5. Then,
E1 = {(1, 4), (2, 3), (3, 2), (4, 1)}
⇒ P(E1) = 4
Therefore, P(E1) = n(E1)/n(S) = 4/36 = 1/9
⇒ odds in favour of E1 = P(E1)/[1 – P(E1)] = (1/9)/(1 – 1/9) = 1/8.

(ii) the odds against getting the sum 6:

Let E2 be the event of getting the sum 6. Then,
E2 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
⇒ P(E2) = 5
Therefore, P(E2) = n(E2)/n(S) = 5/36
⇒ odds against E2 = [1 – P(E2)]/P(E2) = (1 – 5/36)/(5/36) = 31/5.

5. Two dice, one blue and one orange, are rolled simultaneously. Find the probability of getting 

(i) equal numbers on both 

(ii) two numbers appearing on them whose sum is 9.

Solution:

The possible outcomes are 

(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)

What is the probability of getting 2 atleast in one uppermost face when two dice are thrown?

Therefore, total number of possible outcomes = 36.

(i) Number of favourable outcomes for the event E

                   = number of outcomes having equal numbers on both dice 

                   = 6    [namely, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)].

So, by definition, P(E) = \(\frac{6}{36}\)

                                 = \(\frac{1}{6}\)


(ii) Number of favourable outcomes for the event F

           = Number of outcomes in which two numbers appearing on them have the sum 9

            = 4     [namely, (3, 6), (4, 5), (5, 4), (3, 6)].

Thus, by definition, P(F) = \(\frac{4}{36}\)

                                    = \(\frac{1}{9}\).

These examples will help us to solve different types of problems based on probability for rolling two dice.

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