Bahrain
A die is thrown once. Find the probability of getting a multiple of ^@3^@.
Answer:
^@ \dfrac { 1 } { 3 } ^@
- When a die is thrown, the possible outcomes are ^@ 1,2,3,4,5,^@ and ^@6^@.
Total number of possible outcomes = ^@6^@. - Let ^@ E ^@ be the event of getting a multiple of ^@3^@.
Then, the favourable outcomes are getting a ^@ 3 ^@ and ^@ 6 ^@ .
Number of favourable outcomes = ^@2^@. @^ \begin{aligned} \therefore \space P(\text{ getting an a multiple of 3 }) = P(E) = \dfrac { \text { Number of favourable outcomes } } { \text { Total number of outcomes } } = \dfrac { 2 } { 6 } = \dfrac { 1 } { 3 } \end{aligned} @^
