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.
Answer:
1/6
1/6
1/36
Independent
Step-by-step explanation:
Make a chart with 36 outcomes of a pair of six sided dice.
Count the amount that starts with 4, which is 6. 6/36 = 1/6
Doubles. Count the pairs, basically both numbers are the same. 6 outcomes out of total is 1/6.
Last one.
(1/6)^2 = 1/36
These are independent, because the first die landing on 4 doesn't have anything to do with making the die becoming doubles.
