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Sagot :
1) A rolled die has just 6 outcomes; from 1 to 6
[tex]\text{Probability = }\frac{number\text{ of required events}}{nu\text{mber of total events}}[/tex]Number of total events for a die = 6
[tex]\begin{gathered} a)\text{ p(6) } \\ \text{for this the number of required events = 1 because there can and there is only one six showing at a time} \\ p(6)\text{ =}\frac{1}{6} \end{gathered}[/tex][tex]\begin{gathered} b)\text{ p(even)} \\ Here\text{ number of total events are 1,2,3,4,5 and 6} \\ \text{The number of even numbers = 3} \\ \\ p(\text{even) =}\frac{3}{6}=\frac{1}{2} \end{gathered}[/tex][tex]\begin{gathered} c)p(\text{greater than 1)} \\ \text{Here total number of outcomes are 1,2,3,4,5 and 6} \\ \text{numbers greater than 1 are 2,3,4,5 and 6}\ldots..\text{ Th}ere\text{ are 5 of them} \\ \text{Hence} \\ p(\text{greater than 1) =}\frac{5}{6} \end{gathered}[/tex][tex]\begin{gathered} 2)\text{ Total number of cherries = 12} \\ p(\text{sweet) =}\frac{number\text{ of sw}eet\text{ cherries}}{Total\text{number of cherries}} \\ \text{number of swe}et\text{ cherries= 4} \\ p(\text{sweet) =}\frac{4}{12}=\frac{1}{3} \end{gathered}[/tex]
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