91Ó°ÊÓ

$S=(b,y,r)$; b, y, r is the number on the blue, yellow and red die, respectively }

The number of elements in S, namely |S| is$6^3$

We consider two events:

$B<Y<R$The blue ball's value is less than the yellow die's value. which is less than the number of the red die

Compute:

$\text{a)}P\left(D\right)=\text{?}$

localid="1649408084900" $\text{b)}P(B<Y<Râˆ\pounds D)=\text{?}$

$\text{c)}P(B<Y<R)=\text{?}$

For P(D)

Count the number of possible outcomes for the dice. Choose the blue die number six times, the yellow die number five times (but not the blue die number), and any of the four remaining red die numbers.

The probability formula produces for an equally likely set of events:

localid="1649409546971" $$\begin{gathered} \text{a)}P\left(D\right)=\frac{6×5×4}{6^3} \\ =\frac{5}{9} \end{gathered}$$

If the dice all land on the same number, such as D, they can be permuted in a variety of ways on the blue, yellow, and red die. Only one of those six rolls the blue die, the yellow die, and the red die, with the lowest number on the blue die, the middle number on the yellow die, and the highest number on the red die.$B<Y<R.$

$$\begin{gathered} \text{b)}P(B<Y<Râˆ\pounds D) \\ =\frac{1}{6} \end{gathered}$$

First note that:

$B<Y<R⊆D⇒(B<Y<R)âˆ\copyright D=B<Y<R$

We can derive a multiplication rule for two events from the definition of conditional probability.

localid="1649408390569" $$\begin{gathered} P(B<Y<R) \\ =P\left(\right(B<Y<R)âˆ\copyright D) \end{gathered}$$
$=P(B<Y<Râˆ\pounds D)P\left(D\right)$

localid="1649409564502" $=\frac{1}{6}×\frac{5}{9}$

$=\frac{5}{54}$

localid="1649656032093" $≈0.09259$