Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and   red, 4 black balls respectively. O

Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and \lambda  red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola y^{2}=\lambda x with one vertex at the vertex of the parabola, is

Option: 1

432


Option: 2

"__"


Option: 3

"__"


Option: 4

"__"


Answers (1)

best_answer

\begin{aligned} P(\text { Red from } C) & =\frac{\frac{1}{3} \times \frac{\lambda}{\lambda+4}}{\frac{1}{3} \cdot \frac{\lambda}{\lambda+4}+\frac{1}{3} \cdot \frac{4}{10}+\frac{1}{3} \cdot \frac{5}{10}} \\ & =\frac{\frac{\lambda}{\lambda+4}}{\frac{\lambda}{\lambda+4}+\frac{9}{10}} \end{aligned}

\begin{aligned} & \Rightarrow \frac{10 \lambda}{10 \lambda+9(\lambda+4)}=\frac{4}{10} \\ & \Rightarrow 100 \lambda=40 \lambda+36 \lambda+144 \\ & 24 \lambda=144 \\ & \lambda=6 \end{aligned}

\begin{gathered} \mathrm{m}=\frac{2}{t}=\frac{1}{\sqrt{3}} \\ t=2 \sqrt{3} \\ P(12 \mathrm{a}, 4 \sqrt{3} a) \\ (\text { Side })^2=144 a^2+48 a^2 \\ =192 \times \frac{9}{4}=432 \end{gathered}

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question