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for two rolls of a die, the dice fall independently. an even number shows on the first die and the event that an even number shows on the second die?

this is what i got=

the probability that the first die shows an odd number is 1/2, and the probability that the second show an even number is 1/2 . Since the dice fall independently, P(even n odd) = P(first is odd)P(second is even) = (1/2)(1/2) = 1/4.

i'm just wondering, given that the 2 events are independent, then how come is it independent events? because it is conditional that the first number is odd "AND" the second number is even, multiply the two probabilities?

this is for my midterm reviewing~.

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To prove that the events $$ A = \text{"the first number is odd"} $$ and $$B = \text{"the second number is even"} $$ are independent, you have to show that $P(A \cap B) = P(A)P(B)$. But we have $$ A = \{1,3,5\} \times \{1,2,3,4,5,6\},$$

$$ B = \{1,2,3,4,5,6\} \times \{2,4,6\}, $$ and $$ A \cap B = \{1,3,5\} \times \{2,4,6\} $$ That gives $|A| = 18$, $|B| = 18$, $|A \cap B| = 9$, so, as we are in a Laplacian space, $$ P(A) = P(B) = \frac{18}{36} = \frac 12, \quad P(A\cap B) = \frac 9{36} = \frac 14. $$ So $P(A \cap B) = P(A)P(B)$ and $A$ and $B$ are independent.

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    $\begingroup$ if both dice show even numbers? still the same right? the different thing is {1,3,5}*{1,3,5} ,, maybe i should review the laplacian space,,,new to me. $\endgroup$ – kurtk3 Sep 15 '15 at 13:18
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    $\begingroup$ The set $\{1,3,5\} \times \{1,3,5\} = \{(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)\}$, also has nine elements. $\endgroup$ – martini Sep 15 '15 at 13:19

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