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Two dice are thrown simultaneously, what is the probability of getting the same number on both dice?

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

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

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