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MHT CET202014 Oct 2020Evening ShiftMathematicsProbabilityActual

The probability distribution of a discrete r. v. X is array |c|c|c|c|c|c| X =x & 0 & 1 & 2 & 3 & 4 P ( X =x) & k & 2 k & 4 k & 2 k & k array then value of P ( X 2) is

Options

  1. A1 10
  2. B7 10
  3. C3 10
  4. D9 10

Correct answer

B. 7 10

Step-by-step solution

Here k +2 k +4 k +2 k + k =10 k =1 k = 1 10 Now P ( X 2)= P (0)+ P (1)+ P (2)= 1 10 + 2 10 + 4 10 = 7 10

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