Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Parcels from sender receiver <img alt="\mathrm{R}" src="https://entrancecorner.oncodecogs.com/gif.lat

Parcels from sender \mathrm{S} receiver \mathrm{R} pass sequentially through two post-offices. Each post-office has a probability \frac{1}{5} of losing an incoming parcel, independently of all other parcels. Given that a parcel is lost, the probability that it was lost by the second post-office is____.

(enter the answer upto one digit right of demical)

Option: 1

0.4


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Probability to lost at Post Office-1
\mathrm{=P\left(E_{1}\right)=\frac{1}{5}}
 Probability to lost at Post Office-2
\mathrm{=P\left(E_{2}\right)=P\left(E_{1}\right): P\left(E_{2}\right)=\frac{4}{5} \times \frac{1}{5}=\frac{4}{25}}

\mathrm{P(A)=\text{Total Probability to lost}}
             \mathrm{=P\left(E_{1} \cup E_{2}\right)=P\left(E_{1}\right)+P\left(E_{2}\right)}

   (as \mathrm{E_{1}, E_{2}} are mutually exclusive)
 

=\frac{1}{5}+\frac{4}{25}=\frac{9}{25}

\mathrm{\text{Required Probability }=\mathrm{P}\left(\right. Lost\, by \, 2^{\text {nd }} / Parcel \: is \: lost )}
                                                 \mathrm{=P\left(E_{2} / A\right)=\frac{P\left(E_{2} \cap A\right)}{P(A)}=\frac{P\left(E_{2}\right)}{P(A)} }
                                                 \mathrm{=\frac{4 / 25}{9 / 25}=\frac{4}{9}=0.4 } 

Posted by

vishal kumar

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: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question