A die is thrown once. What is the probability of getting a number lying between ^@ 1 ^@ and ^@ 5 ^@ ?


Answer:

^@ \dfrac { 1 } { 2 } ^@

Step by Step Explanation:
  1. We know that in a single throw of a die we can get ^@ 1, 2, 3, 4, 5, ^@ or ^@6.^@
    So, the total number of possible outcomes =^@ 6 ^@.
  2. Let ^@E^@ be an event of getting a number lying between ^@ 1 ^@ and ^@ 5 ^@.
    The event ^@E^@ will happen when the number on the die is ^@ 2, 3, ^@ or ^@ 4 ^@.
    Thus, the number of favorable outcomes = ^@3^@.
  3. We know that the probability of occurrence of an event ^@ E ^@, denoted by ^@P(E)^@ is defined as @^ P(E) = \dfrac { \text { Number of favorable outcomes } } { \text { Total number of possible outcomes } } @^
  4. Therefore, ^@ P (\text{ getting a number lying betweeen } 1 \text{ and } 5 ) = P(E) = \dfrac { 3 } { 6 } = \dfrac { 1 } { 2 } ^@

You can reuse this answer
Creative Commons License

What they say