Find the probability of each outcome when a biased die is rolled, if rolling a 2 or rolling a 4 is three times as likely as rolling each of the other four numbers on the die and it is equally likely to roll a 2 or a 4.

Let's x be the probability for 1, 3, 5 and 6.

The probability for 2 and 4 is going to be 3x.

The sum of the probabilities of all possible outcomes is always 1.

P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1

x + 3x + x + 3x + x + x = 1

10x = 1

x = 1/10

The probability of obtaining 1, 3, 5 or 6 is 1/10

The probability for 2 and 4 is 3/10

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation:

look at photo