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Probability question

2013 · Shift 1 · Q37
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  5. /2013 · Shift 1 · Q37

Probability question

2013 · Shift 1 · Q37

JEE AdvancedMathematicsProbabilityMCQ+4 / −1
Four persons independently solve a certain problem correctly with probabilities 12,34,14,18.{1 \over 2},{3 \over 4},{1 \over 4},{1 \over 8}.21​,43​,41​,81​. Then the probability that the problem is solved correctly by at least one of them is
  1. A
    235256{{235} \over {256}}256235​
  2. B
    21256{{21} \over {256}}25621​
  3. C
    3256{{3} \over {256}}2563​
  4. D
    253256{{253} \over {256}}256253​
View written solutionFree

Correct answer: A

  1. Let the probabilities that the four persons solve the problem correctly be p1=12,p2=34,p3=14,p4=18.p_1=\frac12,\quad p_2=\frac34,\quad p_3=\frac14,\quad p_4=\frac18.p1​=21​,p2​=43​,p3​=41​,p4​=81​.

  2. We need the probability that at least one solves it correctly.

    It is easier to use the complement: P(at least one correct)=1−P(none correct).P(\text{at least one correct})=1-P(\text{none correct}).P(at least one correct)=1−P(none correct).

  3. Since the persons work independently, P(none correct)=(1−p1)(1−p2)(1−p3)(1−p4).P(\text{none correct})=(1-p_1)(1-p_2)(1-p_3)(1-p_4).P(none correct)=(1−p1​)(1−p2​)(1−p3​)(1−p4​).

    Now compute each failure probability: 1−p1=1−12=12,1-p_1=1-\frac12=\frac12,1−p1​=1−21​=21​, 1−p2=1−34=14,1-p_2=1-\frac34=\frac14,1−p2​=1−43​=41​, 1−p3=1−14=34,1-p_3=1-\frac14=\frac34,1−p3​=1−41​=43​, 1−p4=1−18=78.1-p_4=1-\frac18=\frac78.1−p4​=1−81​=87​.

  4. Multiply:

    =\frac{1\cdot1\cdot3\cdot7}{2\cdot4\cdot4\cdot8} =\frac{21}{256}.$$

  5. Therefore, P(at least one correct)=1−21256=256−21256=235256.P(\text{at least one correct})=1-\frac{21}{256}=\frac{256-21}{256}=\frac{235}{256}.P(at least one correct)=1−25621​=256256−21​=256235​.

  6. Comparing with the options, the correct option is A 235256.\boxed{\text{A }\frac{235}{256}}.A 256235​​.

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