Two dice are rolled together find the probability of getting a prime number on each dice

n(s) = 36 i.e.

(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)}

Event = { sum as a prime number }
i.e., = {(1, 1), (1, 2), (2, 1), (1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5) and (5, 6)}

n(E) = 15
P(E) = ?

∴ P(E) = `"n(E)"/"n(s)" = 15/36 = 5/12`

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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.

What is the probability of getting twin prime numbers when two dice are thrown at a time?

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 primes are (2,3) (2, 5) (3, 2) (3, 5) (5, 2) (5, 3) i.e. total 6 pairs

Total outcomes = 36
Favorable outcomes = 6

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

So, P(p,p) = 1/6.

Similar Questions

Question 1: What is the probability of getting even numbers 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 even numbers are (2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6) i.e.  9 pairs

Total outcomes = 36
Favorable outcomes = 9

Probability of getting pair with both even numbers= Favorable outcomes / Total outcomes                                                                                                      = 9 / 36 = 1/4

 So, P(E,E) = 1/4.

Question 2: What is the probability of getting pair with both odd numbers 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 both odd numbers are (1,1) (1,3) (1,5) (3,1) (3,3) (3,5) (5,1) (5,3) (5,5)  i.e. total 9 pairs

Total outcomes = 36
Favorable outcomes = 9

Probability of getting the pair with both odds = Favorable outcomes / Total outcomes                                                                                                       = 9/36 = 1/4

So, P(O,O) = 1/4.

Question 3: What is the probability of getting a pair with one even and one odd number 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 one odd and one even number are (1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (3,6) (4,1) (4,3) (4,5) (5,2) (5,4) (5,6) (6,1) (6,3) (6,5) i.e. total 18 pairs

Total outcomes = 36
Favorable outcomes = 18

Probability of getting a pair with one odd and one even number = Favorable outcomes / Total outcomes                                                                                                      = 18 / 36 = 1/2

So, P(E,O or O,E) = 1/2.

What is the probability of getting a prime number on each dice?

What is the probability of getting two prime numbers?

How many prime numbers are there in two dice?

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