Question

question 2

carl throws a single die twice in a row. for the first throw, carl rolled a 2; for the second throw, he rolled a 4.

what is the probability of rolling a 2 and then a 4?

\\(\frac{1}{3}\\)

\\(\frac{1}{36}\\)

\\(\frac{1}{6}\\)

\\(\frac{1}{12}\\)

Explanation:

Identify the probability of each independent roll

$$
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$$

Calculate the joint probability of both independent events

$$ P(\text{rolling a 2 and then a 4}) = P(\text{rolling a 2}) \times P(\text{rolling a 4}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} $$

Answer:

• (A) \(\frac{1}{3}\)
• (B) \(\frac{1}{36}\) (Correct answer)
• (C) \(\frac{1}{6}\)
• (D) \(\frac{1}{12}\)