Question

A dice is thrown 100 times and the outcomes are noted as given below:

Outcome	1	2	3	4	5	6
Frequency	21	14	18	15	23	9

When a dice is thrown at random, what is the probability of getting a (i) 3, (ii) 6, (iii) 6, (iv) 1?

Answer 1

Total number of trials = 100
Number of times 3 is obtained = 18
Number of times 6 is obtained = 9
Number of times 4 is obtained = 15
Number of times 1 is obtained = 21

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}3=\frac{\mathrm{Number}\mathrm{of}\mathrm{times}3\mathrm{is}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{18}{100}=\frac{9}{50} \\ \left(\mathrm{ii}\right)\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}6=\frac{\mathrm{Number}\mathrm{of}\mathrm{times}6\mathrm{is}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{9}{100} \\ \left(\mathrm{iii}\right)\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}4=\frac{\mathrm{Number}\mathrm{of}\mathrm{times}4\mathrm{is}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{15}{100}=\frac{3}{20} \\ \left(\mathrm{iv}\right)\mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}1=\frac{\mathrm{Number}\mathrm{of}\mathrm{times}1\mathrm{is}\mathrm{obtained}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{21}{100} \end{gathered}$$