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Two number cubes are rolled.

What is the probability that the first lands on 2 and the secend lands on a number greater than 4?

Enter you anwser, as a fraction in simplified form, in the box.

Verified Answer :

rosheerah

(1/6) times (2/6) = 2/36 = 1/18

There is a 1/18 chance.

Answer:

The probability is:

                      \(\dfrac{1}{18}\)

Step-by-step explanation:

We know that the rolling of a first die is independent of the rolling of the second die.

Let A denote the event of rolling a 2 on first die.

and B denote the event of rolling a number greater than 2 on the second die.

and Let P denote the probability of an event.

Also we have:

           P(A)=1/6

and   P(B)=2/6

Hence, the probability that the first land on 2 and the second land on a number greater than 4 i.e. we are asked to find:

                      P(A∩B)

As the events A and B are independent.

Hence we have:

P(A∩B)=P(A)×P(B)

P(A∩B)=2/36

                             P(A∩B)=1/18

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