If a die is thrown at random, then the expectation of the number on it is

If a die is thrown at random, then the expectation of the number on it is
  1. 2.4
  2. 3.5
  3. 2.1
  4. 3.3

Solution

When a die is thrown, probability of getting any number from 1 to 6 is $\frac{1}{6}$. Expectation of a number on it to occur $\begin{array}{l} =\left(1 \times \frac{1}{6}\right)+\left(2 \times \frac{1}{6}\right)+\left(3 \times \frac{1}{6}\right)+\left(4 \times \frac{1}{6}\right)+\left(5 \times \frac{1}{6}\right)+\left(6 \times \frac{1}{6}\right) \\ =\frac{1}{6}(1+2+3+4+5+6)=\frac{21}{6}=3.5 \end{array}$

Asked in: MHT CET 2020 (16 Oct Shift 2)

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