Question

a die is thrown twice find the probability of getting a total of 9​

Answer 1

When a Die is thrown twice...

the total number of outcomes is n(s)=36

The number of possible outcome of getting the sum of total 9 is n(e)=4

Then the probability becomes....

P(e)=n(e)/n(s)

=4/36

=1/9

This the answer

Answer 2

The probability of getting a total of 9 is $\dfrac{1}{9}$.

Step-by-step explanation:

Given : A die is thrown twice.

To find : The probability of getting a total of 9​?

Solution :  

When two dice are thrown,

(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)  

The favorable outcome is getting a total of 9 i.e. {(3,6),(4,5),(5,4),(6,3)}= 4

Total number of outcome = 36

The probability of getting a total of 9 by,

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

$\text{Probability}=\frac{4}{36}$

$\text{Probability}=\frac{1}{9}$

#Learn more  

If two dice are thrown simultaneously, then the probability of getting a doublet or a total of 6 is

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