daily question :) (medium)

Of course, a dice has 6 faces that are all equally likely to occur.

A) Let's make a table that illustrates the probability of just one 6 showing up on one roll:
 

As you can see from this table, there is only a 1/6 probability of getting a 6 the first time. What about three times in a row? Multiply the probabilities of all independent events together. Each event has a 1/6 chance. 

\(P=(\frac{1}{6})^3=\frac{1}{6}*\frac{1}{6}*\frac{1}{6}=\frac{1}{216}\approx0.46\%\)

Therefore, there is a 1/216 chance in rolling a 6 three times in a row. 

B)

We'll use the same method as above. However, these events aren't all the same, so we'll have to calculate the events independently. 

Let's do it for rolling a 5

The probability of rolling a five is alos 1/6. Let's try the next event for seeing the chances of rolling a 1:

Here, too, there is a 1/6 probability of rolling a 1. Let's find the possibility of rolling an even number:

Now that we have determined the probability of each event occurring. Multiply the probabilities of each event together to get the total probability:

\(\frac{1}{6}*\frac{1}{6}*\frac{1}{2}=\frac{1}{72}\approx1.39\%\)

Therefore, there is a 1/72 chance of rolling a 5, then a 1, and then an even number.

C)

Use the same method as illustrated above.

Let's first find the probability of getting an odd number:

Now, let's find the probability of a number greater than 2:

Finally, the last one, which is rolling a 5:

Multiply the probabilities

\(\frac{1}{2}*\frac{2}{3}*\frac{1}{6}=\frac{2}{36}=\frac{1}{18}\approx5.56\%\)

YOu're done now!

A)  Three 6's  =  (1/6)^3  =  1 / 216

B) 5, 1, even   =  (1/6) (1/6) (1/2)  = 1 / 72

C)  Odd, >2, 5  =   (1/2) (4 / 6) (1/6)  =  ( 1/2) (2/3) (1/6)    2 / 36   =   1 / 18