Question
Two dice are thrown simultaneously. What is the probability of getting the sum of the face number is at most 5?
Answer: Option D
In a simultaneous throw of two dice, we have n(s) = 6 × 6 = 36
Let E = event of getting two numbers whose sum is at most 5
Then E = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}
therefore, n(E) = 10
And p(E) = p( getting two numbers whose sum is at most 5)
$$\eqalign{
& {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{{36}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{{18}} \cr} $$
Hence the answer is $$\frac{{5}}{{18}}$$
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In a simultaneous throw of two dice, we have n(s) = 6 × 6 = 36
Let E = event of getting two numbers whose sum is at most 5
Then E = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}
therefore, n(E) = 10
And p(E) = p( getting two numbers whose sum is at most 5)
$$\eqalign{
& {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{{36}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{{18}} \cr} $$
Hence the answer is $$\frac{{5}}{{18}}$$
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