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MHT CET202013 Oct 2020Morning ShiftMathematicsProbabilityActual

Given below is the probability distribution of discrete r.v. X array |c|c|c|c|c|c|c| X =x & 1 & 2 & 3 & 4 & 5 & 6 P [ X =x] & k & 0 & 2 k & 5 k & k & 3 k array Then P [ X 4]=

Options

  1. A1 4
  2. B1 3
  3. C1 2
  4. D3 4

Correct answer

D. 3 4

Step-by-step solution

Here k+0+2 k+5 k+k+3 k=1 k= 1 12 P(X 4)=5 k+k+3 k=9 k= 9 12 = 3 4

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