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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Probability

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Probability

A company has two plants A and B to manufacture motorcycles. 60 \%...

A company has two plants AA and BB to manufacture motorcycles. 60%60 \% motorcycles are manufactured at plant AA and the remaining are manufactured at plant B.80%B .80 \% of the motorcycles manufactured at plant AA are rated of the standard quality, while 90%90 \% of the motorcycles manufactured at plant BB are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If pp is the probability that it was manufactured at plant BB, then 126p126 p is

Options

  1. A54
  2. B66
  3. C64
  4. D56

Correct answer

A. 54

Step-by-step solution

<tableclass="qx−match−listqx−match−table"border="1"cellpadding="6"cellspacing="0"><tr><td></td><td>A</td><td>B</td></tr><tr><td>Manufactured</td><td>60%</td><td>40%</td></tr><tr><td>Standardquality</td><td>80%</td><td>90%</td></tr></table>P<table class="qx-match-list qx-match-table" border="1" cellpadding="6" cellspacing="0"><tr><td></td><td>A</td><td>B</td></tr><tr><td>Manufactured</td><td>60 \%</td><td>40 \%</td></tr><tr><td>Standard quality</td><td>80 \%</td><td>90 \%</td></tr></table> \mathrm{P} (Manufactured at B/\mathrm{B} / found standard quality )=)= ? A : Found S.Q B : Manufacture B C\mathrm{C} : Manufacture A P(E1)=40100P(E2)=60100P(A/E1)=90100P(A/E2)=80100∵P(E1/A)=P(A/E1)P(E1)P(A/E1)P(E1)+P(A/E2)P(E2)=37∴126P=54\begin{aligned} & P\left(E_1\right)=\frac{40}{100} \\ & P\left(E_2\right)=\frac{60}{100} \\ & P\left(A / E_1\right)=\frac{90}{100} \\ & P\left(A / E_2\right)=\frac{80}{100} \\ & \because P\left(E_1 / A\right)=\frac{P\left(A / E_1\right) P\left(E_1\right)}{P\left(A / E_1\right) P\left(E_1\right)+P\left(A / E_2\right) P\left(E_2\right)}=\frac{3}{7} \\ & \therefore 126 P=54 \end{aligned}
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