When two dice are tossed what is the probability of getting 6?

We will be using the concept of the probability of rolling a die to solve this.

Answer: The probability of rolling a sum of 6 with two dice is 5/36.

Let's solve this step by step.

Explanation:

We know, the probability of an event = Favorable outcomes / Total outcomes.

In a case of two dices,

Total outcomes = 6 × 6 = 36

A.T. Q favorable outcome is a sum of 6,

Ways of obtaining a sum of 6

(1,5), (5,1), (2,4),(4,2) and (3,3). Thus, there are 5 ways in which 6 can be obtained using two dices.

Therefore, required probability (P),

P = 5/ 36

Hence, the probability of rolling a sum of 6 with two dice is 5/36.

That's right. The easier approach would be to calculate the chance of not rolling a $6$ - that's just $\frac56$ for the first die, and $\frac56$ for the second die, so by the product rule (as the events are independent), the probability is $\frac56 \cdot \frac56 = \frac{25}{36}$.

Then the probability of rolling a $6$ is $1$ minus the probability of not rolling a $6$, which we just calculated: so it is $1-\frac{25}{36}=\frac{11}{36}$.

If calculating the probability of an event not occurring to calculate the probability of it occurring feels weird to you, you might want to read up on complementary events. The idea here is that the product rule can sometimes make probabilities smaller when that wouldn't make any sense --- in our example, if you had just multiplied $\frac16 \cdot \frac16 = \frac1{36}$, then that would've been obviously wrong. This is because when we work with probability, we're dealing with quantities in the $[0,1]$ interval, so multiplication usually makes things smaller, rather than larger.

Probability means Possibility. It states how likely an event is about to happen. The probability of an event can exist only between 0 and 1 where 0 indicates that event is not going to happen i.e. Impossibility and 1 indicates that it is going to happen for sure i.e. Certainty.

The higher or lesser the probability of an event, the more likely it is that the event will occur or not respectively. For example – An unbiased coin is tossed once. So the total number of outcomes can be 2 only i.e. either “heads” or “tails”. The probability of both outcomes is equal i.e. 50% or 1/2.

So, the probability of an event is Favorable outcomes/Total number of outcomes. It is denoted with the parenthesis i.e. P(Event).

P(Event) = N(Favorable Outcomes) / N (Total Outcomes)

Note: If the probability of occurring of an event A is 1/3 then the probability of not occurring of event A is 1-P(A) i.e. 1- (1/3) = 2/3

What is Sample Space?

All the possible outcomes of an event are called Sample spaces.

Examples-

• A six-faced dice is rolled once. So, total outcomes can be 6 and 
  Sample space will be [1, 2, 3, 4, 5, 6]
• An unbiased coin is tossed, So, total outcomes can be 2 and 
  Sample space will be [Head, Tail]
• If two dice are rolled together then total outcomes will be 36 and 
  Sample space will be  
  (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) ]

Types of Events

Independent Events: If two events (A and B) are independent then their probability will be

P(A and B) = P (A ∩ B) = P(A).P(B) i.e. P(A) * P(B)

Example: If two coins are flipped, then the chance of both being tails is 1/2 * 1/2 = 1/4

Mutually exclusive events:

• If event A and event B can’t occur simultaneously, then they are called mutually exclusive events.
• If two events are mutually exclusive, then the probability of both occurring is denoted as P (A ∩ B) and 
  P (A and B) = P (A ∩ B) = 0
• If two events are mutually exclusive, then the probability of either occurring is denoted as P (A ∪ B) 
  P (A or B) = P (A ∪ B)    
                   = P (A) + P (B) − P (A ∩ B)         
                   = P (A) + P (B) − 0    
                   = P (A) + P (B)

Example: The chance of rolling a 2 or 3 on a six-faced die is P (2 or 3) = P (2) + P (3) = 1/6 + 1/6 = 1/3

Not Mutually exclusive events: If the events are not mutually exclusive then

P (A or B) = P (A ∪ B) = P (A) + P (B) − P (A and B)

What is Conditional Probability?

For the probability of some event A, the occurrence of some other event B is given. It is written as P (A ∣ B)

P (A ∣ B) = P (A ∩ B) / P (B)

Example- In a bag of 3 black balls and 2 yellow balls (5 balls in total), the probability of taking a black ball is 3/5, and to take a second ball, the probability of it being either a black ball or a yellow ball depends on the previously taken out ball. Since, if a black ball was taken, then the probability of picking a black ball again would be 1/4, since only 2 black and 2 yellow balls would have been remaining, if a yellow ball was taken previously, the probability of taking a black ball will be 3/4.

What’s the probability of rolling a sum of 6 on two dice?

Solution:

When two dice are rolled together then total outcomes are 36 and 
Sample space is 
[ (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) ]

So, pairs with sum 6 are (1,5) (2, 4) (3, 3) (4, 2) (5, 1)  i.e. total 5 pairs

Total outcomes = 36
Favorable outcomes = 5

Probability of getting pair with sum 6 = Favorable outcomes / Total outcomes                                                                                                                          = 5 / 36 

So, P(6) = 5/36.

Similar Questions

Question 1: What is the probability of getting 6 on both dice?

Solution:

When two dice are rolled together then total outcomes are 36 and 
Sample space is 
[ (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) ]

So, pairs with both 6 are (6,6) i.e. only 1 pair

Total outcomes = 36
Favorable outcomes = 1

Probability of getting pair with both 6 = Favorable outcomes / Total outcomes                                                                                                                         = 1 / 36 

So, P(6,6) = 1/36

Question 2: What is the probability of getting pair with 6 on only one dice?

Solution:

When two dice are rolled together then total outcomes are 36 and 
Sample space is
[ (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) ]

So, pairs with only one 6 are (1,6) (2,6) (3,6) (4,6) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5)  i.e. total 10 pairs

Total outcomes = 36
Favorable outcomes = 10

Probability of getting the pair with only one 6 = Favorable outcomes / Total outcomes                                                                                                                         = 10/36 = 5/18

So, P(pair with one 6) = 5/18

Question 3: What is the probability of getting a pair with at-least one 6 on two dices?

Solution:

When two dice are rolled together then total outcomes are 36 and 
Sample space is 
[ (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) ]

So, pairs with at-least one 6 are (1,6) (2,6) (3,6) (4,6) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) i.e. 11 pairs

Total outcomes = 36
Favorable outcomes = 11

Probability of getting a pair with at-least one 6 = Favorable outcomes / Total outcomes                                                                                                                         = 11 / 36 

So, P(at-least one 6) = 11/36

What is the probability of rolling 2 dice and getting a sum of 6?

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