Verified Solution • 2025–26 Session moderate

QJEEM24Ma162c: JEE Main 2024 Mathematics - Class 12 JEEM24Ma162c Solution: An integer is chosen at random from the integers . The…

An integer is chosen at random from the integers 1,2,3,…,501,2,3, \ldots, 50 . The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is

Multiple Choice Options

  • A 825\frac{8}{25}
  • B 950\frac{9}{50}
  • C 1425\frac{14}{25}
  • D 2150\frac{21}{50}

Step-by-Step Verified Solution

Given set ={1,2,3,……..50}=\{1,2,3, \ldots \ldots . .50\}

P(A)=\mathrm{P}(\mathrm{A})= Probability that number is multiple of 4

P(B)=\mathrm{P(B)}= Probability that number is multiple of 6

P(C)=\mathrm{P}(\mathrm{C})= Probability that number is multiple of 7

Now,

P(A)=1250,P(B)=850,P(C)=750\mathrm{P}(\mathrm{A})=\frac{12}{50}, \mathrm{P}(\mathrm{B})=\frac{8}{50}, \mathrm{P}(\mathrm{C})=\frac{7}{50}

again

amp;P(A∩B)=450,P(B∩C)=150,P(A∩C)=150amp;P(A∩B∩C)=0\begin{aligned}& \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{50}, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{50}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{50} \\& \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0\end{aligned}

Thus

P(Aamp;∪B∪C)=1250+850+750−450−150−150+0amp;=2150\begin{aligned}P(A & \cup B \cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0 \\& =\frac{21}{50}\end{aligned}

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