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Mathematics · Probability

JEE Advanced 2025 — Paper 1 — Question 16

Three students S1,S2S_{1}, S_{2} and S3S_{3} 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

  1. Option A:

    1336\frac{13}{36}

    Correct
  2. Option B:

    13\frac{1}{3}

  3. Option C:

    1960\frac{19}{60}

  4. Option D:

    14\frac{1}{4}

Answer: A

Step-by-step solution

P(U)=1−P(S1′∩S2′∩S3′)=12\quad \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}

⇒P(S1′∩S2′∩S3′)=12;P(S1′)⋅P(S2′)⋅P(S3′)=12\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}

⇒(1−P(S1))(1−P(S2))(1−P(S3))=12\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}

P(V)=P(S1∩ S2′∩S3′)P(S2′∩S3′)=110\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}

⇒P(S1)⋅P(S2′)P(S3′)=110P(S2′)⋅P(S3′)\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)

⇒P(S1)=110\Rightarrow \mathrm{P}\left(\mathrm{S}_{1}\right)=\frac{1}{10}

P(W)=P(S2∩ S3′)=112\mathrm{P}(\mathrm{W})=\mathrm{P}\left(\mathrm{S}_{2} \cap \mathrm{~S}_{3}^{\prime}\right)=\frac{1}{12}

P(S2)⋅P(S3′)=112\mathrm{P}\left(\mathrm{S}_{2}\right) \cdot \mathrm{P}\left(\mathrm{S}_{3}^{\prime}\right)=\frac{1}{12}

P(S2)(1−P(S3))=112\mathrm{P}\left(\mathrm{S}_{2}\right)\left(1-\mathrm{P}\left(\mathrm{S}_{3}\right)\right)=\frac{1}{12}

Eq. (1) (1−110)(1−P(S2))(1−P(S3))=12\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}

(1−P(S2))(1−P(S3))=59\left(1-\mathrm{P}\left(\mathrm{S}_{2}\right)\right)\left(1-\mathrm{P}\left(\mathrm{S}_{3}\right)\right)=\frac{5}{9}

 Eq.(2)  Eq.(3) ⇒P(S2)1−P(S2)=112×95\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} P(S2)=323\mathrm{P}\left(\mathrm{S}_{2}\right)=\frac{3}{23}

Put in Eq. (2) 323(1−P(S3))=112\frac{3}{23}\left(1-P\left(S_{3}\right)\right)=\frac{1}{12}

1−P(S3)=23361-\mathrm{P}\left(\mathrm{S}_{3}\right)=\frac{23}{36}

P(S3)=1336P\left(S_{3}\right)=\frac{13}{36}

P(T)=1336\mathrm{P}(\mathrm{T})=\frac{13}{36}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2025
Paper
Paper 1
Subject
Mathematics
Chapter
Probability
Topic
Conditional Probability and Multiplication Theorem