What is the probability of a die landing on a 3 the first toss, a 1 the second toss, and then a 4 on the third toss? Dice- 6 sides

Please help

  2087568668/1  

step-by-step explanation:

the answer is ! ?

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116