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If the events A and B are mutually exclusive events such that P(A)= 1 3 (3 x+1) and P(B)= 1 4 (1-x) then the possible values of x lies in the interval

Options

  1. A[- 7 9 , 4 9 ]
  2. B[- 1 3 , 5 9 ]
  3. C[0,1]
  4. D[ 1 3 , 2 9 ]

Correct answer

B. [- 1 3 , 5 9 ]

Step-by-step solution

For any event E , the probability P(E) must satisfy 0 P(E) 1 . For event A , 0 1 3 (3x+1) 1 . Multiplying by 3 , 0 3x+1 3 . Subtracting 1 , -1 3x 2 . Dividing by 3 , - 1 3 x 2 3 . For event B , 0 1 4 (1-x) 1 . Multiplying by 4 , 0 1-x 4 . Subtracting 1 , -1 -x 3 . Multiplying by -1 , -3 x 1 . Since A and B are mutually exclusive, P(A B) = P(A) + P(B) 1 . 3x+1 3 + 1-x 4 1 . Multiplying by 12 , 4(3x+1) + 3(1-x) 12 . 12x + 4 + 3 - 3x 12 . 9x + 7 12 . 9x 5 . x 5 9 . Also, P(A) 0 and P(B) 0 implies x -1/3 and x 1 . Comb

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