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Mathematics · Probability

JEE Advanced 2019 — Paper 1 — Question 43

Let SS be the sample space of all 3×33 \times 3 matrices with entries from the set {0,1}\{0,1\}. Let the events E1\mathrm{E}_{1} and E2\mathrm{E}_{2} be given by E1={A∈S:det⁡A=0} and E2={A∈S: sum of entries of A is 7}\begin{aligned} & E_{1}=\{A \in S: \operatorname{det} A=0\} \text { and } & E_{2}=\{A \in S: \text { sum of entries of } A \text { is } 7\} \end{aligned} If a matrix is chosen at random from SS, then the conditional probability P(E1/E2)P\left(E_{1} / E_{2}\right) equals ____\_\_\_\_

Answer: 0.5

Numerical answer — enter this value.

Step-by-step solution

n(E2)=n\left(E_{2}\right)= arrangement of 7,1 and 2 or =9!7!2!=36=\frac{9!}{7!2!}=36

n(E1∩E2)=\mathrm{n}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right)= both zero should be in a row or a column

=[111111100]\begin{aligned} & =\left[\begin{array}{lll} 1 & 1 & 1 \\1 & 1 & 1\\ 1 & 0 & 0 \end{array}\right] \end{aligned}

number of ways of arranging of (1,0,0)=3(1,0,0)=3 and arrangement of row=3=3

total =9=9

in same way for (1,0,0)(1,0,0) for columns number of ways will be =9=9

total ways =18=18

P(E1E2)=P(E1∩E2)P(E2)=1836=12=0.50P\left(\frac{E_{1}}{E_{2}}\right)=\frac{P\left(E_{1} \cap E_{2}\right)}{P\left(E_{2}\right)}=\frac{18}{36}=\frac{1}{2}=0.50

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2019
Paper
Paper 1
Subject
Mathematics
Chapter
Probability
Topic
Conditional Probability and Multiplication Theorem

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