Problem Solving

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

IMO C

plz explain the logic

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]

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

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

prob of getting any odd number is 2/3 how come ? i find its 1/3

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.