KCET · Maths · Mathematical Reasoning
If two dice are thrown simultaneously, then the probability that the sum of the numbers which
come up on the dice to be more than \( 5 \) is
- A \( \frac{5}{36} \)
- B \( \frac{1}{6} \)
- C \( \frac{5}{18} \)
- D \( \frac{13}{18} \)
Answer & Solution
Correct Answer
(D) \( \frac{13}{18} \)
Step-by-step Solution
Detailed explanation
Total number of outcomes is
\(6 \times 6=6\)
Favourable outcomes is \((1,5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6),(4,2),(4,3),(4,4),(4,5),(4,6),(5,\),
\(1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\), that is 26 .
Therefore, required probability is
\(P=\frac{26}{36}=\frac{13}{18}\)
\(6 \times 6=6\)
Favourable outcomes is \((1,5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6),(4,2),(4,3),(4,4),(4,5),(4,6),(5,\),
\(1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\), that is 26 .
Therefore, required probability is
\(P=\frac{26}{36}=\frac{13}{18}\)
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