Question

I roll two dice and get a sum more than 9. what is the probability that the number on the first die is even

Answer 1

Hey!

Favorable outcome possibility={(4,6),(5,6),(6,6),(6,5),(6,4)}=5.
Total possibility=36 as there are two dices.

Probability of sum more than 9 = 5/36.

Hope it helps.

Answer 2

Answer:

To ensure that the first die's number is even, a $\frac{2}{3}$ probability is needed.

Step-by-step explanation:

Let B, represent the scenario in which the first die's number is even.

Let A, be the occurrence of receiving an amount greater than 9.

Than

Therefore, A = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6,6)}

B = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

        (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

        (6, 1), (6, 2), (6,3), (6, 4), (6, 5), (6,6)}

Now,

Total number of outcomes= $6^{2}$=36=n(S)

A∩ B= {(4,6),(6,4),(6,5),(6.6)}

∴n(A)=6;n(B)=18;n(A∩B)=4

∴P(A)=$\frac{n(A)}{n(S)}=\frac{6}{36}$  P(A∩B)=$\frac{n(A n B)}{n(S)}=\frac{4}{36\\}$

Thus,

The necessary probability,

In terms of conditional probability,

$P(\frac{B}{A} )=\frac{P(B n A)}{P(A)} =\frac{\frac{4}{36} }{6} =\frac{2}{3}$)

∴$P(\frac{B}{A} )=\frac{2}{3}$

To learn more about Conditional Probability, visit:

https://brainly.in/question/4102547  

To learn more about First die, visit:

https://brainly.in/question/1478568  

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