Question

You roll a six-sided die twice. What is the probability of rolling a 2 and then an odd number?
A1/18
B 1/6
C1/3
D1/12

Answer 1

answer : D1/12

steps:

'and' in a word problem means multiply

rolling a 2 : 1/6

odd :

1 2 3 4 5 6

1 3 5 are 3 odd numbers so

3/6 which is 1/2

1/6 * 1/2 = 1/12

Answer 2

Answer:

$\textsf{D)} \quad \dfrac{1}{12}$

Step-by-step explanation:

Probability is a measure of how likely events are to happen.

When all the possible outcomes are equally likely, we can use this formula to calculate the probability of an event happening:

$\boxed{\sf P(event)=\dfrac{\textsf{Number of outcomes where event happens}}{\textsf{Total number of possible outcomes}}}$

There are 6 sides on the dice numbered 1 through 6.  

Therefore, the total number of possible outcomes is 6.

Assuming the dice is fair, the probability of rolling a 2 is:

$\sf P(x=2)=\dfrac{1}{6}$

There are 3 odd numbers in the set of numbers {1, 2, 3, 4, 5, 6}.

Therefore, the probability of rolling an odd number is:

$\sf P(x=1)\;or\;P(x=3)\;or\;P(x=5)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}$

Therefore, the probability of rolling a 2 and then an odd number is:

$\sf P(x=2)\;and\;P(x=odd)=\dfrac{1}{6} \times \dfrac{1}{2}=\dfrac{1}{12}$