Question

6. Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.​

Answer 1

Step-by-step explanation:

(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)

Sample space = 36

1.even number on first die

n(A) =18

P(A)=18/36

=2

2.total of face sum of 8

n(B)=5

P(B) =5/36

Answer 2

The probability of getting an even number on the first die or a total of face sum 8 is  $\dfrac{1}{12}$.

Step-by-step explanation:

Given : Two dice are rolled once.

To find : The probability of getting an even number on the first die or a total of face sum 8?

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)  

Event of getting an even number on the first die is {(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)  (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) }

Event of getting a total of face sum 8 is {(2,6) (3,5) (4,4) (5,3) (6,2)}

The favorable outcome of getting an even number on the first die or a total of face sum 8 i.e. {(2,6),(4,4),(6,2)}= 3

Total number of outcome = 36

The probability of getting an even number on the first die or a total of face sum 8 is given by,

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

$\text{Probability}=\frac{3}{36}$

$\text{Probability}=\frac{1}{12}$

#Learn more  

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