Mathematics · Probability

JEE Advanced 2023 — Paper 2 — Question 10

Let $X$ be the set of all five digit numbers formed using 1, 2, 2, 2, 4, 4, 0. For example, 22240 is in $X$ while 02244 and 44422 are not in $X$ . Suppose that each element of $X$ has an equal chance of being chosen. Let $p$ be the conditional probability that an element chosen at random is a multiple of 20 given that it is a multiple of 5 . Then the value of $38 p$ is equal to

Answer: 31

Numerical answer — enter this value.

Step-by-step solution

$0,1,2,2,2,4,4$ Number divisible by 5 (event A) $=38$ Number divisible by 20 and 5(event B) $=31$ $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\mathrm{P}(\mathrm{B}) \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)$ $\Rightarrow \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{31}{38}$

$\therefore 38 \mathrm{p}=31$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2023
Paper: Paper 2
Subject: Mathematics
Chapter: Probability
Topic: Conditional Probability and Multiplication Theorem

Answer: 31

Step-by-step solution

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