What is the probability of rolling a sum less than or equal to 4?

Using our GCF Calculator, we can reduce the top and bottom of this fraction by a greatest common factor (GCF) of 2 to get:

In other words, you have a 72.22% chance (13 out of 18) of rolling greater than or equal to 6

You have a 72.22% chance (13 out of 18) of rolling greater than or equal to 6

Calculates the probability for the following events in a pair of fair dice rolls: * Probability of any sum from (2-12) * Probability of the sum being less than, less than or equal to, greater than, or greater than or equal to (2-12) * The sum being even * The sum being odd * The sum being a prime number * The sum being a non-prime number * Rolling a list of numbers i.e. (2,5,6,12) * Simulate (n) Monte Carlo two die simulations. 2 dice calculator

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2 dice rolldiceobjects used in games of chance wih 6 sidesmonte carlo simulationa model used to predict the probability of a variety of outcomes when the potential for random variables is present.numberan arithmetical value, expressed by a word, symbol, or figure, representing a particular quantity and used in counting and making calculations and for showing order in a series or for identification. A quantity or amount.prime numbera natural number greater than 1 that is not a product of two smaller natural numbers.probabilitythe likelihood of an event happening. This value is always between 0 and 1.
P(Event Happening) = Number of Ways the Even Can Happen / Total Number of Outcomes

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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.

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) ]

P(Sum < 9 ) = P(1) + P(2) + P(3)…… + P(8)

So, pairs with sum less than 9 are (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2,3) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (4,1) (4,2) (4,3) (4,4) (5,1) (5,2) (5,3) (6,1) (6,2) i.e. total 26 pairs

Total outcomes = 36
Favorable outcomes = 26

Probability of getting the sum less than 9 = Favorable outcomes / Total outcomes

                                                                 = 26 / 36 = 13/18

So, P(sum < 9) = 13/18

Similar Questions

Question 1: When two dice are thrown what is the probability that the sum of numbers appeared on them would be less than 5?

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) ]

P(Sum < 5 ) = P(1) + P(2) + P(3) + P(4)

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

Total outcomes = 36
Favorable outcomes = 6

Probability of getting the sum less than 5 = Favorable outcomes / Total outcomes
                                                                 = 6 / 36 = 1/6

So, P(sum < 5) = 1/6

Question 2: When two dice are thrown what is the probability that the sum of numbers appeared on them would be more than 9?

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) ]

Maximum sum possible is 12.

P(Sum > 9 ) = P(10) + P(11) + P(12) 

So, pairs with sum more than 9 are (4,6) (5,5) (5,6) (6,4) (6,5) (6,6) i.e. total 6 pairs

Total outcomes = 36
Favorable outcomes = 6

Probability of getting the sum more than 9 = Favorable outcomes / Total outcomes                                                                                                                           = 6/36 = 1/6

So, P(sum>9) = 1/6

Question 3: When two dice are thrown what is the probability that the sum of numbers appeared on them would be less than 4?

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) ]

Minimum sum possible is 2.

P(Sum < 4 ) = P(2) + P(3) 

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

Total outcomes = 36
Favorable outcomes = 3

Probability of getting the sum less than 4 = Favorable outcomes / Total outcomes                                                                                                                          = 3 / 36 = 1/12

So, P(sum < 4) = 1/12