wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

n letters to each of which corresponds an addressed envelope are placed in the envelopes at random. What is the probability that no letter is placed in the right envelope?

A
1{11!12!+13!+(1)n.1n!}
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
{11!12!+13!+(1)n.1n!}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
{11!+12!+13!++1n!}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1{11!+12!+13!++1n!}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D 1{11!12!+13!+(1)n.1n!}
Let Ath denote the event that the ith letter is placed in the right envelope.
Then the required probability is P(¯A1¯A2.....¯An)P(AiA2......An).
[By De-Morgan law]
=1P(A1A2....An)
=1[P(Ai)p(AiAj)]+P(AiajAk)ijijk
+(1)n1p(AiA2Ak)
Now p(Ai)=(n1)!n! as having placed ith letter in the right envelope, the remaining letters can be placed in (n1)! ways.
Similarly, P(A1A2.....Ar)= Prob.of r particular letters in right envelopes =(nr)!n!
p(A1A2.....Ar)
=nCr.(nr)!n!=1r! where r=1,2,3,....n.
(¯A1¯A2¯An)
=1{11!12!+13!+(1)n.1n!}
which is equal to first n2 terms in the expansion of e1.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon