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Probability question

2019 · 9 Apr · Shift 1 · Q36
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Probability question

2019 · 9 Apr · Shift 1 · Q36

JEE MainMathematicsProbabilityMCQ+4 / −1
Four persons can hit a target correctly with probabilities 12{1 \over 2}21​, 13{1 \over 3}31​, 14{1 \over 4}41​ and 18{1 \over 8}81​ respectively. if all hit at the target independently, then the probability that the target would be hit, is :
  1. A
    2532{{25} \over {32}}3225​
  2. B
    25192{{25} \over {192}}19225​
  3. C
    1192{{1} \over {192}}1921​
  4. D
    732{{7} \over {32}}327​
View written solutionFree

Correct answer: A

  1. Let the probabilities that the four persons hit the target be p1=12,p2=13,p3=14,p4=18.p_1=\frac12,\quad p_2=\frac13,\quad p_3=\frac14,\quad p_4=\frac18.p1​=21​,p2​=31​,p3​=41​,p4​=81​.

  2. We need the probability that the target is hit by at least one person.

  3. It is easier to first find the probability that none of them hits the target.

    Their miss probabilities are: 1−p1=12,1−p2=23,1−p3=34,1−p4=78.1-p_1=\frac12,\quad 1-p_2=\frac23,\quad 1-p_3=\frac34,\quad 1-p_4=\frac78.1−p1​=21​,1−p2​=32​,1−p3​=43​,1−p4​=87​.

  4. Since all hit independently, the probability that all miss is 12⋅23⋅34⋅78.\frac12\cdot\frac23\cdot\frac34\cdot\frac78.21​⋅32​⋅43​⋅87​.

    Simplifying, 12⋅23=13,\frac12\cdot\frac23=\frac13,21​⋅32​=31​, 13⋅34=14,\frac13\cdot\frac34=\frac14,31​⋅43​=41​, 14⋅78=732.\frac14\cdot\frac78=\frac{7}{32}.41​⋅87​=327​.

    So, P(none hits)=732.P(\text{none hits})=\frac{7}{32}.P(none hits)=327​.

  5. Therefore, P(target is hit)=1−P(none hits)=1−732=2532.P(\text{target is hit})=1-P(\text{none hits})=1-\frac{7}{32}=\frac{25}{32}.P(target is hit)=1−P(none hits)=1−327​=3225​.

  6. Now compare with the options:

    • A: 2532\frac{25}{32}3225​
    • B: 25192\frac{25}{192}19225​
    • C: 1192\frac{1}{192}1921​
    • D: 732\frac{7}{32}327​

    Hence the correct option is: A\boxed{A}A​

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