Question

A die is thrown twice and the sum of numbers appearing is observed to be 7. What is the conditional probability that number 2 has appeared at least once.

Answer 1

your probability for atleast 2 is 11/36

Answer 2

Answer:

$\text{Probability}=\frac{11}{36}$

Step-by-step explanation:

Given : A die is thrown twice and the sum of numbers appearing is observed to be 7.

To find : What is the conditional probability that number 2 has appeared at least once.

Solution :  

When two dice rolled once the outcome 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)  

Now,  

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

Favorable outcome that number 2 has appeared at least once :

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (1,2) (3,2) (4,2) (5,2) (6,2)

So, Favorable outcome = 11

Total number of outcome = 36

$\text{Probability}=\frac{11}{36}$