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QJEEM22Ma717d: JEE Main 2022 Mathematics - Class 12 JEEM22Ma717d Solution: Bag A contains 2 white, 1 black and 3 red balls and bag B…

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is 611{6 \over {11}} , then n is equal to __________.

Multiple Choice Options

  • A 13
  • B 6
  • C 4
  • D 3

Step-by-Step Verified Solution

JEE Main 2022 (Online) 24th June Morning Shift Mathematics - Probability Question 83 English Explanation
amp;P(1R and 1B)=P(A)⋅P(1R1BA)+P(B)⋅P(1R1BB)amp;=12⋅3C1⋅1C16C2+12⋅2C1⋅3C1n+5C2amp;P(1R1BA)=12⋅31512⋅315+12⋅6⋅2(n+5)(n+4)=611amp;⇒110110+6(n+5)(n+4)=611\begin{aligned}&P(1 R \text { and } 1 B)=P(A) \cdot P\left(\frac{1 R 1 B}{A}\right)+P(B) \cdot P\left(\frac{1 R 1 B}{B}\right) \\\\&=\frac{1}{2} \cdot \frac{{ }^{3} C_{1} \cdot{ }^{1} C_{1}}{{ }^{6} C_{2}}+\frac{1}{2} \cdot \frac{{ }^{2} C_{1} \cdot{ }^{3} C_{1}}{{ }^{n+5} C_{2}} \\\\&P\left(\frac{1 R 1 B}{A}\right)=\frac{\frac{1}{2} \cdot \frac{3}{15}}{\frac{1}{2} \cdot \frac{3}{15}+\frac{1}{2} \cdot \frac{6 \cdot 2}{(n+5)(n+4)}}=\frac{6}{11} \\\\&\Rightarrow \frac{\frac{1}{10}}{\frac{1}{10}+\frac{6}{(n+5)(n+4)}}=\frac{6}{11}\end{aligned}

amp;⇒1110=610+36(n+5)(n+4)amp;⇒510×36=1(n+5)(n+4)amp;⇒n2+9n−52=0amp;⇒n=4 is only possible value \begin{aligned}&\Rightarrow \frac{11}{10}=\frac{6}{10}+\frac{36}{(n+5)(n+4)} \\\\&\Rightarrow \frac{5}{10 \times 36}=\frac{1}{(n+5)(n+4)} \\\\&\Rightarrow n^{2}+9 n-52=0 \\\\&\Rightarrow n=4 \text { is only possible value }\end{aligned}

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