Suppose that Marvin rolls a pair of fair six-sided dice. Let A be the event that the first die lands on a 4 and B be the event that rolls doubles
P(A), the probability that the first die lands on 4?
P(B), the probability that Marvin rolls doubles?
P(A and B) the probability that the first die lands on 4 and Marvinv rolls doubles?
DEPENDENT OR INDEPENDENT?

The probability that the first die lands on 4 is 1/6.

The probability that Marvin rolls doubles is 1/6.

The probability that the first die lands on 4 and Marvin rolls doubles is 11/36.

The two events A and B are independent.

Step-by-step explanation :

It is given that,

Marvin rolls a pair of fair six-sided dice.

The total number of outcomes when a pair of dice are rolled ⇒ 6² = 36

Let A be the event that the first die lands on a 4.

The possibility to land 4 on the first die are (4,1) (4,2) (4,3) (4,4) (4,5) and (4,6).

∴ P(A) = no.of outcomes with 4 on 1st die / Total no.of outcomes

⇒ 6 / 36

⇒ 1/6

Let B be the event that rolls doubles (a double means that both dice show the same number).

Hence the doubles are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

∴ P(B) = no.of outcomes with doubles / Total no.of outcomes

⇒ 6 / 36

⇒ 1/6

Now, to find P(A and B) the probability that the first die lands on 4 and Marvin rolls double is P(A∪B).

⇒ P(A∪B) = P(A) + P(B) - P(A∩B)

(A∩B) is (4,4) that occurs in both the event A and event B.

∴ P(A∩B) = 1/36

P(A∪B) = 1/6 + 1/6 - 1/36 ⇒ 11/36

To test whether two events A and B are independent, calculate P(A), P(B), and P(A ∩ B).

And then check whether P(A ∩ B) equals P(A)P(B).

If they are equal, A and B are independent; if not, they are dependent.

⇒ P(A)×P(B) = 1/6 × 1/6

⇒ 1/36 which is equal to P(A∩B)

∴ The two events A and B are independent.

its not b

step-by-step explanation: