wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

In a game, a man wins Rs. 1000 if he gets an even number 4 on a fair die and loses Rs. 200 for getting any other number on the die. If he decides to throw the die until he wins or maximum of three times, then his expected gain/loss (in Rupees) is

A
38009 loss
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
38009 gain
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
400 gain
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 38009 gain
Even number on a die 4 is 4,6
Let A be the event that either 4 or 6 appears on the die
and B be the event that a number other than 4,6 appears on the die.
Then P(A)=13 and P(B)=23

Possible sequence of events are
A,BA,BBA,BBB

Let X be a random variable denoting the winning amount.
X+1000+800+600600P(X)1329427827

E(X)=1000×13+800×29+600×427600×827 =38009

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