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QJEEAdv13Maa5ff: JEE Advanced 2013 Mathematics - Class 12 JEEAdv13Maa5ff Solution: Four persons independently solve a certain problem correctly…

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}. Then the probability that the problem is solved correctly by at least one of them is

Multiple Choice Options

  • A 235256{{235} \over {256}}
  • B 21256{{21} \over {256}}
  • C 3256{{3} \over {256}}
  • D 253256{{253} \over {256}}

Step-by-Step Verified Solution

Let A, B, C, D are the events of solving a problem.

Given: P(A)=12,P(B)=34,P(C)=14,P(D)=18\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{3}{4}, \mathrm{P}(\mathrm{C})=\frac{1}{4}, \mathrm{P}(\mathrm{D})=\frac{1}{8}

Also given A, B, C, D are independent events.

Let P is the probability of solved by at least one person.

⇒P=1−\Rightarrow \mathrm{P}=1- (Probability that none of the person can solve the problem)

amp;⇒P=1−P(A‾∩B‾∩C‾∩D‾)amp;⇒P=1−P(A‾)⋅P(B‾)⋅P(C‾)⋅P(D‾)amp;⇒P=1−(1−P(A))(1−P(B))(1−P(C))(1−P(D))amp;⇒P=1−(1−12)(1−34)(1−14)(1−18)amp;⇒P=1−12×14×34×78amp;⇒P=1−21256amp;⇒P=235256\begin{aligned}& \Rightarrow \mathrm{P}=1-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}} \cap \overline{\mathrm{D}}) \\& \Rightarrow \mathrm{P}=1-\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\overline{\mathrm{B}}) \cdot \mathrm{P}(\overline{\mathrm{C}}) \cdot \mathrm{P}(\overline{\mathrm{D}}) \\& \Rightarrow \mathrm{P}=1-(1-\mathrm{P}(\mathrm{A}))(1-\mathrm{P}(\mathrm{B}))(1-\mathrm{P}(\mathrm{C}))(1-\mathrm{P}(\mathrm{D})) \\& \Rightarrow \mathrm{P}=1-\left(1-\frac{1}{2}\right)\left(1-\frac{3}{4}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right) \\& \Rightarrow \mathrm{P}=1-\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} \times \frac{7}{8} \\& \Rightarrow \mathrm{P}=1-\frac{21}{256}\\& \Rightarrow \mathrm{P}=\frac{235}{256}\end{aligned}

Hints:

⇒\Rightarrow (i) If AA and BB are independent events, then

amp;→P(A∩B)=P(A)⋅P(B)amp;→P(A‾∩B‾)=P(A‾)⋅P(B‾)=(1−P(A))(1−P(B))amp;→P(A‾∩B)=P(A‾)⋅P(B)=(1−P(A))P(B)amp;→P(A∩B‾)=P(A)⋅P(B‾)=P(A)⋅(1−P(B))\begin{aligned}& \rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \\& \rightarrow \mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\overline{\mathrm{B}})=(1-\mathrm{P}(\mathrm{A}))(1-\mathrm{P}(\mathrm{B})) \\& \rightarrow \mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})=\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\mathrm{B})=(1-\mathrm{P}(\mathrm{A})) \mathrm{P}(\mathrm{B}) \\& \rightarrow \mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\overline{\mathrm{B}})=\mathrm{P}(\mathrm{A}) \cdot(1-\mathrm{P}(\mathrm{B}))\end{aligned}

(ii) Probability of at least one person can solve the problem =1−= 1- Probability that none of the persons can solve that problem.

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