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AP EAMCET201923 Apr 2019Morning ShiftMathematicsProbabilityActual

If (E₁ ) and (E₂ ) are two events of a random experiment such that (P (E₁ )= 1 8 , P (E₁ E₂ )= 1 3 ), (P (E₂ E₁ )= 1 4 ), then match the items of List-I with the items of List-II. ( array lllc & List-I & & List-II (A) & P (E₂ ) & I. & 3 16 (B) & P (E₁ E₂ ) & II. & 3 29 (C) & P ( E ₁ E ₂ ) & III. & 3 32 (D) & P (E₁ E ₂ ) & IV. & 26 29 & & V. & 13 32 array ) The correct match is

Options

  1. A( array cc & A & B & C & D & I & III & IV & II array )
  2. B( array cc & A & B & C & D & III & I & IV & V array )
  3. C( array cc & A & B & C & D & III & I & IV & II array )
  4. D( array cc & A & B & C & D & I & II & V & IV array )

Correct answer

C. ( array cc & A & B & C & D & III & I & IV & II array )

Step-by-step solution

For two given events (E₁ ) and (E₂ ), the given information are (P (E₁ )= 1 8 , P (E₁ E₂ )= 1 3 ) and ( aligned & P (E₂ E₁ )= 1 4 & P (E₂ E₁ )= 1 4 P (E₁ E₂ ) P (E₁ ) = 1 4 & P (E₁ E₂ )= 1 32 & P (E₂ )= P (E₁ E₂ ) P (E₁ E₂ ) = 1 32 1 3 = 3 32 & P (E₁ E₂ )=P (E₁ )+P (E₂ )-P (E₁ E₂ ) & = 1 8 + 3 32 - 1 32 = 3 16 aligned ) ( aligned P ( E ₁ ) & = 7 8 and P ( E ₂ )= 29 32 and P ( E ₁ E ₂ ) & =P ( E₁ E₂ ) & =1-P (E₁ E₂ )= 13 16 P ( E ₁ E ₂ ) & = P ( E ₁ E ₂ ) P ( E ₂ ) = 13 16 29 32 = 26 29 and P (E₁ E ₂ ) & =1-P ( E ₁

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