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Probability question

2003 · Shift 0 · Q88
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  5. /2003 · Shift 0 · Q88

Probability question

2003 · Shift 0 · Q88

JEE MainMathematicsProbabilityMCQ+4 / −1
Events A,B,CA, B, CA,B,C are mutually exclusive events such that P(A)=3x+13,P(B)=1−x4P\left( A \right) = {{3x + 1} \over 3},P\left( B \right) = {{1 - x} \over 4}P(A)=33x+1​,P(B)=41−x​ and P(C)=1−2x2P\left( C \right) = {{1 - 2x} \over 2}P(C)=21−2x​ The set of possible values of xxx are in the interval.
  1. A
    [0,1]\left[ {0,1} \right][0,1]
  2. B
    [13,12]\left[ {{1 \over 3},{1 \over 2}} \right][31​,21​]
  3. C
    [13,23]\left[ {{1 \over 3},{2 \over 3}} \right][31​,32​]
  4. D
    [1 3,133]\left[ {{1 \ 3},{13 \over 3}} \right][1 3,313​]
View written solutionFree

Correct answer: B

  1. Since A,B,CA,B,CA,B,C are events, each probability must satisfy 0≤P(A),P(B),P(C)≤1.0 \le P(A),P(B),P(C) \le 1.0≤P(A),P(B),P(C)≤1. Also, because they are mutually exclusive, their pairwise intersections are empty, so P(A∪B∪C)=P(A)+P(B)+P(C)≤1.P(A\cup B\cup C)=P(A)+P(B)+P(C)\le 1.P(A∪B∪C)=P(A)+P(B)+P(C)≤1.

  2. Write the given probabilities: P(A)=3x+13,P(B)=1−x4,P(C)=1−2x2.P(A)=\frac{3x+1}{3},\qquad P(B)=\frac{1-x}{4},\qquad P(C)=\frac{1-2x}{2}.P(A)=33x+1​,P(B)=41−x​,P(C)=21−2x​.

  3. Use non-negativity of each probability.

  • For P(A)≥0P(A)\ge 0P(A)≥0: 3x+13≥0  ⟹  3x+1≥0  ⟹  x≥−13.\frac{3x+1}{3}\ge 0 \implies 3x+1\ge 0 \implies x\ge -\frac13.33x+1​≥0⟹3x+1≥0⟹x≥−31​.

  • For P(B)≥0P(B)\ge 0P(B)≥0: 1−x4≥0  ⟹  1−x≥0  ⟹  x≤1.\frac{1-x}{4}\ge 0 \implies 1-x\ge 0 \implies x\le 1.41−x​≥0⟹1−x≥0⟹x≤1.

  • For P(C)≥0P(C)\ge 0P(C)≥0: 1−2x2≥0  ⟹  1−2x≥0  ⟹  x≤12.\frac{1-2x}{2}\ge 0 \implies 1-2x\ge 0 \implies x\le \frac12.21−2x​≥0⟹1−2x≥0⟹x≤21​.

Combining these: −13≤x≤12.-\frac13\le x\le \frac12.−31​≤x≤21​.

  1. Now use mutual exclusiveness: P(A)+P(B)+P(C)≤1.P(A)+P(B)+P(C)\le 1.P(A)+P(B)+P(C)≤1. Substitute: 3x+13+1−x4+1−2x2≤1.\frac{3x+1}{3}+\frac{1-x}{4}+\frac{1-2x}{2}\le 1.33x+1​+41−x​+21−2x​≤1. Take LCM 121212: 4(3x+1)+3(1−x)+6(1−2x)12≤1.\frac{4(3x+1)+3(1-x)+6(1-2x)}{12}\le 1.124(3x+1)+3(1−x)+6(1−2x)​≤1. Simplify numerator: 12x+4+3−3x+6−12x=13−3x.12x+4+3-3x+6-12x = 13-3x.12x+4+3−3x+6−12x=13−3x. So, 13−3x12≤1.\frac{13-3x}{12}\le 1.1213−3x​≤1. Hence, 13−3x≤12  ⟹  −3x≤−1  ⟹  x≥13.13-3x\le 12 \implies -3x\le -1 \implies x\ge \frac13.13−3x≤12⟹−3x≤−1⟹x≥31​.

  2. Combine all conditions: x≥13,x≤12.x\ge \frac13,\qquad x\le \frac12.x≥31​,x≤21​. Therefore, x∈[13,12].x\in \left[\frac13,\frac12\right].x∈[31​,21​].

  3. Match with the options: [13,12]\left[\frac13,\frac12\right][31​,21​] is Option B.

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