Answer: [tex]Pr(rolling\text{ a 5 on the 1st row or an odd number on the 2nd roll\rparen = 0.667}[/tex]
Explanation:
Given:
A die is rolled twice
To find:
the probability of showing 5 on the 1st row and an odd number on the 2nd roll
Possible outcone of a die = {1, 2, 3, 4, 5, 6}
The total number from a die = 6
Pr(rolling a 5) = number of times 5 occurs/total number
number o times 5 occurs = 1
Pr(rolling a 5) = 1/6
Pr(rolling an odd number) = number of odd numbers/total
Odd numbers = {1, 3, 5}
number of odd numbers = 3
Pr(rolling an odd number) = 3/6
the probability (5 on the 1st row or an odd number on the 2nd roll = Pr(rolling a 5) + Pr(rolling an odd number)
[tex]\begin{gathered} Pr(rolling\text{ a 5 or an odd number\rparen = }\frac{1}{6}\text{ + }\frac{3}{6} \\ \\ Pr(rolling\text{ a 5 or an odd number\rparen= }\frac{4}{6}\text{ = 2/3} \\ \\ Pr(rolling\text{ a 5 or an odd number\rparen= 0.667} \\ \end{gathered}[/tex]
