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

A and B take turns in throwing two dice, the first to throw 10 being awarded the prize, show that if A has the first throw, their chance of winning are in the ratio 12 : 11.

Open in App
Solution

There are only three possible cases, wherein the sum of the numbers obtained after throwing 2 dice is 10, i.e. 4, 6 5, 5 6, 4. Psum of the numbers is 10=336=112Psum of the numbers is not 10=1-112=1112Pany numner other than six=56PA winning=P10 in first throw+P10 in third throw+ ...=112+1112×1112×112+ ...=1121+11122+11124+ ... =11211-121144 1+a+a2+a3+ ... =11-a=112×14423=1223PB winning=1-PA winning=1123Now,PA winningPB winning=12231123=1211Hence proved.

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Events
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon