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 a sum of 7 when two dice are thrown?

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 7 are (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) i.e. total 6 pairs

Total outcomes = 36
Favorable outcomes = 6

Probability of getting the sum of 7 = Favorable outcomes / Total outcomes

                                                      = 6 / 36 = 1/6

So, P(sum of 7) = 1/6.

Related Questions

Question 1: What is the probability of getting 1 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 1’s are (1,1) i.e. only 1 pair

Total outcomes = 36
Favorable outcomes = 6

Probability of getting pair of 1 = Favorable outcomes / Total outcomes 
                                                 = 1 / 36 

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

Question 2: What is the probability of getting the sum of 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) ]

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

Total outcomes = 36
Favorable outcomes = 3

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

So, P(sum of 3) = 1/12.

Question 3: What is the probability of getting the sum of 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) ]

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

Total outcomes = 36
Favorable outcomes = 4

Probability of getting the sum of 5 = Favorable outcomes / Total outcomes
                                                       = 4 / 36 = 1/9

So, P(5) = 1/9.