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A single die is rolled twice. The 36​ equally-likely outcomes are shown to the right. Find the probability of getting first an odd number and second an even number.

Respuesta :

jushmk
36 equally-likely outcome: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1),(6,2), (6,3), (6,4), (6,5), (6,6)

Solution:
Outcomes with first number being old number and second being even number: (1,2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2), (5,4), (5,6) = 9 outcomes

P(old,even) = 9/36 =1/4 = 0.25
jscuervo1992

Answer:

The probability of getting first an odd number and second an even number IS 0.25

Step-by-step explanation:

Lets define the following events:

A: Probability of obtaining an odd number first time

B: Probability of obtaining an even number second time

and probability of ane events is defined as

P (Event) = number of ways event may happen / number of possible outcomes

Then:

P(A) = 3/ 6 =1/2

because we can have 1, 3,5 so event A occurs and there are six possible outcomes (1,2,3,4,5,6)

P(B) = 3/ 6  =1/2

because we can have 2, 4,6 so event b occurs and there are six possible outcomes (1,2,3,4,5,6)

the probaility that both events occur one after the other is

P(A and B) = P(A)*P(B) = 1/2* 1/2 = 1/4 = 0.25

This applies only if the chances that event B happening doesnt change after A happens. As A event is independent from B event, we can do this multiplication.

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