2 marksare three fair siz-sided dice. Each die has two faces, two yellow faces and two red faces. All three dice are rolled. Find the probability of rolling exactly one red face

14.81%

Step-by-step explanation:

A fair six-sided dice has a total of six sides meaning that there are 6 possible outcomes for each dice roll. If two of the faces on the dice are red, then the possibility of rolling a red on a single dice roll would be 2/6, while the probability of rolling anything else would be 4/6 because there are 4 other sides left. Now if we roll three seperate dice, in order to calculate the probability of getting only one red face we need to multiply the probability of getting a red face on the first dice by the probability of the other two dice getting anything else.

2/6 * 4/6 * 4/6 = 32/216 or 0.1481

We can multiply this decimal by 100 to get the percentage value...

0.1481 * 100 = 14.81%

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}