Question
Two dice are thrown simultaneously. What is the probability of getting the sum of the face number is odd?
Answer: Option A
In a simultaneous throw of two dice, we have n(S) = 6 x 6 = 36
Let E = event of getting two numbers whose sum is odd.
Then E = {(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)}
therefore, n(E) = 18
And P(E) = p( getting two numbers whose sum is odd)
$$\eqalign{
& {\text{P}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{18}}{{36}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr} $$
Hence the answer is $$\frac{1}{2}$$
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In a simultaneous throw of two dice, we have n(S) = 6 x 6 = 36
Let E = event of getting two numbers whose sum is odd.
Then E = {(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)}
therefore, n(E) = 18
And P(E) = p( getting two numbers whose sum is odd)
$$\eqalign{
& {\text{P}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{18}}{{36}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr} $$
Hence the answer is $$\frac{1}{2}$$
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