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MHT CET202613 April 2026Evening ShiftMathematicsProbabilityActual

It is known that a box of 8 batteries contains 3 defective pieces and a person randomly selects 2 batteries from this box. Then the probability distribution of the number of defective batteries is

Options

  1. AX = x 0 1 2 P(X = x) 10 28 15 28 3 28
  2. BX = x 1 2 3 P(X = x) 10 28 15 28 3 28
  3. CX = x 0 1 2 P(X = x) 15 28 10 28 3 28
  4. DX = x 1 2 3 P(X = x) 15 28 10 28 3 28

Correct answer

A. X = x 0 1 2 P(X = x) 10 28 15 28 3 28

Step-by-step solution

Total number of batteries = 8 Number of defective batteries = 3 Number of non-defective batteries = 5 Let X be the random variable representing the number of defective batteries selected. Since 2 batteries are drawn, the possible values of X are 0, 1, 2 . Total number of ways to select 2 batteries from 8 is ⁸C₂ = 28 . Probability of selecting 0 defective batteries ( X = 0 ): P(X = 0) = ⁵C₂ ⁸C₂ = 10 28 Probability of selecting 1 defective battery ( X = 1 ): P(X = 1) = ³C₁ ⁵C₁ ⁸C₂ = 3 5 28 = 15 28 Probability of sele

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