Let two fair six-faced dice A and B be thrown | vedclass

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(B) The sample space $S$ has $6 	imes 6 = 36$ outcomes.$E_1 = \{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$,so $P(E_1) = \frac{6}{36} = \frac{1

Vedclass · Accepted Answer

$E_1, E_2$ and $E_3$ are independent.

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(B) The sample space $S$ has $6 	imes 6 = 36$ outcomes.$E_1 = \{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$,so $P(E_1) = \frac{6}{36} = \frac{1}{6}$.$E_2 = \{(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)\}$,so $P(E_2) = \frac{6}{36} = \frac{1}{6}$.$E_3$ is the event that the sum is odd,which occurs if one die is even and the other is odd. $P(E_3) = \frac{18}{36} = \frac{1}{2}$.$E_1 \cap E_2 = \{(4, 2)\}$,so $P(E_1 \cap E_2) = \frac{1}{36} = P(E_1)P(E_2)$. Thus,$E_1$ and $E_2$ are independent.$E_1 \cap E_3 = \{(4, 1), (4, 3), (4, 5)\}$,so $P(E_1 \cap E_3) = \frac{3}{36} = \frac{1}{12} = P(E_1)P(E_3)$. Thus,$E_1$ and $E_3$ are independent.$E_2 \cap E_3 = \{(1, 2), (3, 2), (5, 2)\}$,so $P(E_2 \cap E_3) = \frac{3}{36} = \frac{1}{12} = P(E_2)P(E_3)$. Thus,$E_2$ and $E_3$ are independent.For $E_1, E_2, E_3$ to be independent,we must have $P(E_1 \cap E_2 \cap E_3) = P(E_1)P(E_2)P(E_3)$.$E_1 \cap E_2 \cap E_3 = \{(4, 2)\} \cap E_3 = \emptyset$ because the sum $4+2=6$ is even. So $P(E_1 \cap E_2 \cap E_3) = 0$.Since $0 
eq \frac{1}{6} 	imes \frac{1}{6} 	imes \frac{1}{2} = \frac{1}{72}$,the events are not mutually independent.

Let two fair six-faced dice $A$ and $B$ be thrown simultaneously. If $E_1$ is the event that die $A$ shows up $4$,$E_2$ is the event that die $B$ shows up $2$,and $E_3$ is the event that the sum of numbers on both dice is odd,then which of the following statements is $NOT$ true?

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