Question

numbers 1-20 were writtenon chits of paper and dropped in box what is the probability of

a-choosing muliples of 10

b-choosing an even number

c-choosing an odd prime number

d-choosing a factor of 6

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{sample}\mathrm{space}\mathrm{is}\mathrm{given}\mathrm{by}: \\ \mathrm{S}=\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\right\} \\ \mathrm{Thus}n\left(\mathrm{S}\right)=20 \\ \underline{\left(\mathbf{a}\right).}\mathrm{Let}{\mathrm{E}}_1=\mathrm{event}\mathrm{of}\mathrm{choosing}\mathrm{a}\mathrm{multiple}\mathrm{of}10=\left\{10,20\right\} \\ \mathrm{Then},n\left({\mathrm{E}}_1\right)=2 \\ \mathrm{Therefore},p\left({\mathrm{E}}_1\right)=\frac{n\left({\mathrm{E}}_1\right)}{n\left(\mathrm{S}\right)}=\frac{2}{20}=\frac{1}{10} \\ \underline{\left(\mathbf{b}\right).}\mathrm{Let}{\mathrm{E}}_2=\mathrm{event}\mathrm{of}\mathrm{choosing}\mathrm{an}\mathrm{even}\mathrm{number}=\left\{2,4,6,8,10,12,14,16,18,20\right\} \\ \mathrm{Then},n\left({\mathrm{E}}_2\right)=10 \\ \mathrm{Therefore},\mathit{}p\left({\mathrm{E}}_2\right)=\frac{n\left({\mathrm{E}}_2\right)}{n\left(\mathrm{S}\right)}=\frac{10}{20}=\frac{1}{2} \end{gathered}$$


$$\begin{gathered} \underline{\left(\mathbf{c}\right)\mathbf{.}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathrm{Let}{\mathrm{E}}_3=\mathrm{event}\mathrm{of}\mathrm{choosing}\mathrm{an}\mathrm{odd}\mathrm{prime}\mathrm{number}=\left\{3,5,7,11,13,17,19\right\} \\ \mathrm{Then},n\left({\mathrm{E}}_3\right)=7 \\ \mathrm{Therefore},\mathit{}p\left({\mathrm{E}}_3\right)=\frac{n\left({\mathrm{E}}_3\right)}{n\left(\mathrm{S}\right)}=\frac{7}{20} \\ \underline{\left(\mathbf{d}\right)\mathbf{.}}\mathrm{Let}{\mathrm{E}}_4=\mathrm{event}\mathrm{of}\mathrm{choosing}\mathrm{a}\mathrm{factor}\mathrm{of}6=\left\{1,2,3,6\right\} \\ \mathrm{Then},n\left({\mathrm{E}}_4\right)=4 \\ \mathrm{Therefore},\mathit{}p\left({\mathrm{E}}_4\right)=\frac{n\left({\mathrm{E}}_4\right)}{n\left(\mathrm{S}\right)}=\frac{4}{20}=\frac{1}{5} \end{gathered}$$