Suppose that a fair, 6 sided die is rolled. Let X indicate the event that an even number is rolled (in other words, X = 1 if an even number is rolled and X = 0 otherwise). Let Y indicate the event that 2, 3, or 4 is rolled (in other words, Y = 1 if 2, 3, or 4 is rolled and Y = 0 otherwise). Find P (X = 0, Y = 1).

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uchedan uchedan · Answer 1 · 2020-03-08T02:44:40+01:00

Answer: 1/2, 1/2

Step-by-step explanation:

A die has 6 faces.

Total number of faces = 6

Possible outcomes are :1,2,3,4,5,6

Even numbers : 2, 4, and 6 = 3

Non even numbers: 1,3,and 5= 3

First statement:

P(Even number) = 3/ 6 = 1/2

P(X=1) = 1/2

P(odd number) = 3/6 = 1/2

P(X=0) = 1/2

Second statement:

P(2) = 1/6

P(3) = 1/6

P(4) = 1/6

P(2 or 3 or 4) = (1/6) + (1/6) + (1/6)

P(2 or 3 or 4) = (1+1+1)/6 = 3/6 =1/2

Thus,

P(Y=1) = 1/2

Therefore,

P(X =0,Y=1) = 1/2, 1/2

lhmarianateixeira lhmarianateixeira · Answer 2 · 2022-03-08T11:20:02+01:00

In this exercise we have to use the knowledge of probability to calculate the chance of an event to occur, so:

[tex]P(X =0,Y=1) = 1/2, 1/2[/tex]

organizing the following information given in the text we have that:

Then calculating the probability we find that:

[tex]P(Even number) = 3/ 6 = 1/2\\P(X=1) = 1/2\\P(odd number) = 3/6 = 1/2\\P(X=0) = 1/2[/tex]

For the second case the probability will be:

[tex]P(2) = 1/6\\P(3) = 1/6\\P(4) = 1/6\\P(2 or 3 or 4) = (1/6) + (1/6) + (1/6)\\P(2 or 3 or 4) = (1+1+1)/6 = 3/6 =1/2[/tex]

See more about probability at brainly.com/question/795909