BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

defective die

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 232
Joined: Fri Jun 18, 2010 7:09 am
Thanked: 1 times
Followed by:2 members

defective die

by ruplun » Fri Jul 29, 2011 5:20 am
In a defective six-sided dice the probability of getting an odd number is twice the probability of getting an even number.What is the probability of getting 5 in a single throw?

a. 1/18
b. 1/9
c. 2/9
d. 1/2
e. 2/3
Top
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 60
Joined: Sat Jul 16, 2011 11:56 pm
Location: hyderabad
Thanked: 2 times
Followed by:1 members

by naveen451 » Fri Jul 29, 2011 9:05 am
IMO C
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 232
Joined: Fri Jun 18, 2010 7:09 am
Thanked: 1 times
Followed by:2 members

by ruplun » Fri Jul 29, 2011 9:05 pm
plz explain the logic
Top
Senior | Next Rank: 100 Posts
Posts: 34
Joined: Thu Jun 30, 2011 4:15 am
Thanked: 1 times
Followed by:1 members

by ntamhane » Fri Jul 29, 2011 9:29 pm
What's the OA?

My Logic:
In a non defective or unbiased dice the Probability of getting a odd number or a even number is 1/2
The question suggests it is a six sided defective dice with the probability of getting a odd number is 2 times getting the probability of an even number.
Hence the numbers on the dice will be
odd odd odd
even even odd
In this case the Probability of getting the odd number is 4/6 = 2/3
Probability of getting the even number is = 2/6 = 1/3
which satisfies the condition "probability of getting a odd number is 2 times getting the probability of an even number"
Now the extra odd number could be 5 or any other odd number.
i.e
1 3 5
2 4 5
OR
1 3 5
2 4 3(or any other)
hence the probability of getting a 5 in a single throw is 2/6 or 1/6
[spoiler]2/6 + 1/6 = 3/6 = 1/2[/spoiler]
Top
Senior | Next Rank: 100 Posts
Posts: 54
Joined: Wed Jul 20, 2011 12:36 pm
Thanked: 2 times

by sumasajja » Sat Jul 30, 2011 3:27 am
actually given here that the probability of throwing odd numbers is 2
so in a dice for numbers 1 2 3 4 5 6
the sample space is 2+1+2+1+2+1=9
and for getting 5 on dice the defective probability is 2
so probability of getting '5' is
=no.of outcomes /sample space
=2/9
Top
User avatar
Legendary Member
Posts: 1255
Joined: Fri Nov 07, 2008 2:08 pm
Location: St. Louis
Thanked: 312 times
Followed by:90 members

by Tani » Sat Jul 30, 2011 4:33 am
ntamhane -

Your analysis starts correctly. The probability of getting an odd number is indeed 2/3. But, there are three odd numbers so the probability of getting any specific odd number is 1/3 of the probability of getting any odd number, so (2/3)/3 = 2/9
Tani Wolff
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 232
Joined: Fri Jun 18, 2010 7:09 am
Thanked: 1 times
Followed by:2 members

by ruplun » Sat Jul 30, 2011 8:41 am
prob of getting any odd number is 2/3 how come ? i find its 1/3
Top
User avatar
Legendary Member
Posts: 1255
Joined: Fri Nov 07, 2008 2:08 pm
Location: St. Louis
Thanked: 312 times
Followed by:90 members

by Tani » Sat Jul 30, 2011 7:43 pm
The probability of getting an odd number is twice the probability of getting an even number. Together they must total 1. So the probability of odd is 2/3 and the probability of even is 1/3.
Tani Wolff
Top
Post new topic Post Reply
• Page 1 of 1