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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Mathematics
Chapter
Probability

JEE Advanced 2025 · Paper 1 · Mathematics · Probability

Three students S_1, S_2 and S_1 are given a problem to solve....

Three students S1,S2S_1, S_2 and S1S_1 are given a problem to solve. Consider the following events : U:U: At least one of S1,S2S_1, S_2 and S3S_3 can solve the problem, V:S1V: S_1 can solve the problem, given that neither S2\mathrm{S}_2 nor S3\mathrm{S}_3 can solve the problem, W:S2W: S_2 can solve the problem and S3S_3 cannot solve the problem, T:S3T: S_3 can solve the problem. For any event EE, let P(E)P(E) denote the probability of EE. If P(U)=12,P(V)=110P(U)=\frac{1}{2}, P(V)=\frac{1}{10} and P(W)=112P(W)=\frac{1}{12}, then P(T)P(T) is equal to

Options

  1. A1336\frac{13}{36}
  2. B13\frac{1}{3}
  3. C1960\frac{19}{60}
  4. D14\frac{1}{4}

Correct answer

A. 1336\frac{13}{36}

Step-by-step solution

P(U)=1−P(S1′∩S2′∩S3′)=12⇒P(S1′∩S2′∩S3′)=12;P(S1′)⋅P(S2′)⋅P(S3′)=12⇒(1−P(S1))(1−P(S2))(1−P(S3))=12…(1)\begin{aligned} & \mathrm{P}(\mathrm{U})=1-\mathrm{P}\left(\mathrm{ S}_1^{\prime} \cap \mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)=\frac{1}{2} \\ & \Rightarrow \mathrm{P}\left(\mathrm{ S}_1^{\prime} \cap \mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)=\frac{1}{2} ; \mathrm{P}\left(\mathrm{ S}_1^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{ S}_2^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{ S}_3^{\prime}\right)=\frac{1}{2} \\ & \Rightarrow\left(1-\mathrm{P}\left(\mathrm{ S}_1\right)\right)\left(1-\mathrm{P}\left(\mathrm{ S}_2\right)\right)\left(1-\mathrm{P}\left(\mathrm{ S}_3\right)\right)=\frac{1}{2} \quad \ldots(1) \end{aligned}P(V)=P(S1∩S2′∩S3′)P(S2′∩S3′)=110⇒P(S1)⋅P(S2′)P(S3′)=110P(S2′)⋅P(S3′)⇒P(S1)=110\begin{aligned} & \mathrm{P}(\mathrm{V})=\frac{\mathrm{P}\left(\mathrm{S}_1 \cap \mathrm{ S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)}{\mathrm{P}\left(\mathrm{S}_2^{\prime} \cap \mathrm{S}_3^{\prime}\right)}=\frac{1}{10} \\ & \Rightarrow \mathrm{P}\left(\mathrm{S}_1\right) \cdot \mathrm{P}\left(\mathrm{S}_2^{\prime}\right) \mathrm{P}\left(\mathrm{S}_3^{\prime}\right)=\frac{1}{10} \mathrm{P}\left(\mathrm{S}_2^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{S}_3^{\prime}\right) \\ & \Rightarrow \mathrm{P}\left(\mathrm{S}_1\right)=\frac{1}{10}\end{aligned} P(W)=P(S2∩S3′)=112P(S2)⋅P(S3′)=112P(S2)(1−P(S3))=112...(2)\begin{aligned} & \mathrm{P}(\mathrm{ W})=\mathrm{P}\left(\mathrm{ S}_2 \cap \mathrm{ S}_3^{\prime}\right)=\frac{1}{12} \\ & \mathrm{P}\left(\mathrm{ S}_2\right) \cdot \mathrm{P}\left(\mathrm{ S}_3^{\prime}\right)=\frac{1}{12} \\ & \mathrm{P}\left(\mathrm{ S}_2\right)\left(1-\mathrm{P}\left(\mathrm{ S}_3\right)\right)=\frac{1}{12}...(2) \end{aligned}Eq.(1)Eq. (1)\left(1-\frac{1}{10}\right)\left(1-\mathrm{P}\left(\mathrm{ S}_2\right)\right)\left(1-\mathrm{P}\left(\mathrm{ S}_3\right)\right)=\frac{1}{2} \left(1-\mathrm{P}\left(\mathrm{ S}_2\right)\right)\left(1-\mathrm{P}\left(\mathrm{ S}_3\right)\right)=\frac{5}{9}....(3)  Eq. (2) Eq. (3)⇒P(S2)1−P(S2)=112×95P(S2)=323\begin{aligned} & \frac{\text{ Eq. }(2)}{\text{ Eq. }(3)} \Rightarrow \frac{\mathrm{P}\left(\mathrm{ S}_2\right)}{1-\mathrm{P}\left(\mathrm{ S}_2\right)}=\frac{1}{12} \times \frac{9}{5} \\ & \mathrm{P}\left(\mathrm{ S}_2\right)=\frac{3}{23} \end{aligned}PutinEq.(2)Put in Eq. (2)323(1−P(S3))=1121−P(S3)=2336P(S3)=1336P(T)=1336\begin{aligned} & \frac{3}{23}\left(1-\mathrm{P}\left(\mathrm{ S}_3\right)\right)=\frac{1}{12} \\ & 1-\mathrm{P}\left(\mathrm{ S}_3\right)=\frac{23}{36} \\ & \mathrm{P}\left(\mathrm{ S}_3\right)=\frac{13}{36} \\ & \mathrm{P}(\mathrm{ T})=\frac{13}{36} \end{aligned}$
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