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