The die can land in 6 different ways.
Only one of them is a 2.
So the probability of rolling a 2 on any roll is 1/6 .
In a large number of rolls, you expect roughly 1/6 of them to be a 2.

In 108 rolls, you should expect a 2 somewhere around

(108/6) = 18 times.

3 0

4 years ago

Read 2 more answers

Can you please help me out

Taya2010 [7]

The bag contains,

Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,

Total marbles (possible outcome) is,

$\text{Total marbles = (R) + (G) +(B) = 9 + 7 + 4 = 20 marbles}$

Let P(R) represent the probablity of picking a red marble,

P(G) represent the probability of picking a green marble and,

P(B) represent the probability of picking a blue marble.

Probability , P, is,

$\text{Prob, P =}\frac{required\text{ outcome}}{possible\text{ outcome}}$ $\begin{gathered} P(R)=\frac{9}{20} \\ P(G)=\frac{7}{20} \\ P(B)=\frac{4}{20} \end{gathered}$

Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,

That means once a marble is drawn, the total marbles (possible outcome) reduces as well,

$\begin{gathered} \text{Prob of a red marble P(R) =}\frac{9}{20} \\ \text{Prob of }a\text{ blue marble =}\frac{4}{19} \\ \text{After a marble is selected without replacement, marbles left is 19} \\ \text{Prob of red marble + prob of blue marble = P(R) + P(B) = }\frac{9}{20}+\frac{4}{19}=\frac{251}{380} \\ \text{Hence, the probability is }\frac{251}{380} \end{gathered}$

Hence, the best option is G.

5 0

2 years ago

Which expression is equivalent to?...

stiks02 [169]

Answer:

B x^9 ∛ y

Step-by-step explanation:

(x^27 y) ^ (1/3)

x^27^(1/3) y^(1/3)

We can distribute the exponent a^b ^c = a^(b*c)

x^(27*1/3) y^(1/3)

x^9 y^(1/3)

3 0

4 years ago