A die is thrown twice. What is the probability that the number in the second throw is higher than the number in the first throw ?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\frac{5}{12}\)
Two throws give \(6\times6=36\) equally likely outcomes. In 6 of them the two numbers are equal, and the other 30 split evenly between “second higher” and “first higher”, because each such pair can be read either way round.
Equal numbers: (1, 1), (2, 2), …, (6, 6) \(=6\) outcomes
Second higher \(=\displaystyle\frac{36-6}{2}=15\) outcomes
Probability \(=\displaystyle\frac{15}{36}=\frac{5}{12}\)