Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Cards are drawn one by one at random from a well-shuffled full pack of 52 playing cards until two aces are obtained for the first time. If <img alt="\mathrm{N}" src="https://entrancecorner.oncodeco

Cards are drawn one by one at random from a well-shuffled full pack of 52 playing cards until two aces are obtained for the first time. If \mathrm{N} is the number of cards required to be drawn then \mathrm{P(N=n)=\frac{(n-1)(\lambda-n)(\mu-n)}{\alpha \times \beta \times \gamma \times \delta}}where \mathrm{2 \leq n \leq 50}, then the value of \mathrm{\alpha+\beta+\gamma+\delta+2 \lambda+3 \mu} must be

Option: 1

385


Option: 2

387


Option: 3

386


Option: 4

\frac{1}{4}


Answers (1)

best_answer

If \mathrm{n} is the number of draws, then \mathrm{2 \leq n \leq 50}, since \mathrm{nth} draw must be an ace and one ace must appear in first \mathrm{(n-1)} draws. Let \mathrm{S} be the sample space.

Then, \mathrm{n(S)=} Total number of ways of drawing the first \mathrm{(n-1)} cards out of 52 .
\mathrm{ ={ }^{52} C_{n-1} }

and let \mathrm{E } be the event of favourable cases.

\mathrm{\therefore n(E)= }number of ways of choosing 1 ace in \mathrm{(n-1) }
trials \times number of ways for the \mathrm{n}th draw

\mathrm{ =\left({ }^4 C_1 \times{ }^{48} C_{n-2}\right) \times \frac{3}{52-(n-1)} }

{\because number of favourable ways is 3}

\mathrm{ =4 \times \frac{48 !}{(n-2) !(50-n) !} \times \frac{3}{(53-n)}}

Hence, the required probability, \mathrm{ P(E)=\frac{n(E)}{n(S)}}

\mathrm{=\frac{\frac{4 \times 481 \times 3}{(n-2) ! \times(50-n) ! \times(54-n)}}{^{52}C_{n-1}} }

\mathrm{=\frac{\frac{4 \times 481 \times 3}{(n-2) ! \times(50-n) ! \times(53-n)}}{\frac{52 !}{(n-1) !(53-n) !}} }

\mathrm{=\frac{12 \times 481 \times(n-1) ! \times(53-n) !}{521 \times(n-2) ! \times(50-n) \mid \times(53-n)} }

\mathrm{=\frac{12 \times 481 \times(n-1)(n-2) ! \times(53-n)(52-n) !}{52 \times 51 \times 50 \times 49 \times 48 ! \times(n-2) ! \times(50-n) ! \times(53-n)} }

\mathrm{ =\frac{12 \times(n-1) \times(52-n) !}{52 \times 51 \times 50 \times 49 \times(50-n) !} }

\mathrm{ =\frac{(n-1)(52-n)(51-n)(50-n) !}{50 \times 49 \times 17 \times 13 \times(50-n) !} }

\mathrm{ =\frac{(n-1)(52-n)(51-n)}{50 \times 49 \times 17 \times 13} }

\mathrm{ \therefore \quad \lambda=52, \mu=51, \alpha=50, \beta=49, \gamma=17, \delta=13 }

\mathrm{ \therefore \alpha+\beta+\gamma+\delta+2 \lambda+3 \mu=50+49+17+13+104+153 }

                                                      \mathrm{ =386 }

Hence option 3 is correct.

 









 

Posted by

manish

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 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question