A group of pirates captures Kevin, Lisa, Matt and Neal, and forces them to play a game. They each roll a fair 6-sided-die once. If the product of their roll is a multiple of 3, they all have to walk the plank, but otherwise they are safe. What is the probability that they survive? A)2/3 B)16/81 C)145/1296 D)65/81 E)625/1296 PLZ answer been waiting. I'll give 30 points

Option B, 16/81

Step-by-step explanation:

So we have 4 prisoners, they will roll a fair six side die and the product of the four rolls must NOT be a multiple of 3.

We know that every integer number can be "decomposed" into a product of prime numbers.

Then a number N,  that is divisible by 3, can be written as:

N = 3*k

Where k is another integer.

Here we will have a product of 4 numbers, each of them are in between 1 and 6.

Now, if only one of the prisoners rolls a 3, then the product of the rolls will always be a multiple of 3. And if one of the rolls is 6 the same will happen, because 6 = 3.2

Then the probability of surviving is when in none of the four rolls we have a 3 or a 6.

Then we must have a 1, 2, 4 or 5.

The probability of 4 outcomes out of 6, is:

P = 4/6.

But we have 4 rolls, so we have that probability four times, and the joint probability will be equal to the product of the probabiliities for each roll, then the probability of surviving is:

P = (4/6)^4 = (2/3)^4 = 16/81

-22

times both sides by 2, times () by 8, add alile terms, move terms, add alike and caculate divide both sides by 3