Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
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]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
