Question

if a die is rolled twice ,find the probability of getting an even number in the first time or a total of 8

Answer 1

Answer:

Step-by-step explanation:

Total possible outcomes 36

Probability of getting even number in first time 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)

Probability of getting 8 on total(2,6)(3,5),(4,4)(5,3)(6,2)

Total outcomes=22-2=20

Probability=20/36=5/9

Hope it helps

Answer 2

The probability of getting an even number in the first time or a total of 8 is $\frac{1}{12}$

Step-by-step explanation:

Given : If a die is rolled twice.

To find : The probability of getting an even number in the first time or a total of 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)  

The even number in the first time are (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) .

Outcome of total of 8 are (2,6),(3,5),(4,4),(5,3),(6,2)

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

Total number of outcome = 36

The probability of getting an even number in the first time or a total of 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  

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

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