Probability Problem

This topic has expert replies
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Fri Jan 07, 2011 9:17 pm

Probability Problem

by nehatandon » Fri Jan 07, 2011 9:23 pm
A certain dice game can only be won if, when a player throws two fair six-sided dice, the number showing on one of the dice
is a multiple of the number showing on the other. What is the probability that a player wins this game?
(A)11/18
(B)11/36
(C)1/4
(D)2/9
(E)7/36
Source: — Problem Solving |

User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pm
Thanked: 147 times
Followed by:3 members

by anshumishra » Fri Jan 07, 2011 9:44 pm
nehatandon wrote:A certain dice game can only be won if, when a player throws two fair six-sided dice, the number showing on one of the dice
is a multiple of the number showing on the other. What is the probability that a player wins this game?
(A)11/18
(B)11/36
(C)1/4
(D)2/9
(E)7/36
Here are the ways in which a player can win

1st dice------------ 2nd dice
1------------ 1,2,3,4,5,6
2------------ 1,2,4,6
3------------ 1,3,6
4------------ 1,2,4
5------------ 1,5
6------------ 1,2,3,6

Favorable no. of outcomes = 6+4+3+3+2+4 = 22
Total no. of outcomes = 6*6 = 36

So, required probability = 22/36 = 11/18 A
Thanks
Anshu

(Every mistake is a lesson learned )

User avatar
Senior | Next Rank: 100 Posts
Posts: 67
Joined: Fri Nov 26, 2010 8:25 am
Thanked: 3 times

by Ramit88 » Fri Jan 07, 2011 11:32 pm
is one a multiple of every number?

User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pm
Thanked: 147 times
Followed by:3 members

by anshumishra » Fri Jan 07, 2011 11:35 pm
Ramit88 wrote:is one a multiple of every number?
Nope, but every no. is multiple of one. If you get (say) no. 2 on the 1st dice and no. 1 on 2nd dice, still you win. (As the number showing on one of the dice is still a multiple of number showing on the other).
Thanks
Anshu

(Every mistake is a lesson learned )