Counterfeit dice roll question

• Apr 15th 2010, 11:30 AM
chrisceville22
Counterfeit dice roll question
A six sided die, faces numbered 1 - 6 as usual, is known to be counterfeit. The probability of rolling an even number is twice of the probability of rollong an odd number. What is the probability that, if the die is throw twice, the first roll will be a 5 and the second will be a 6?

A) 2/8! B) 1/18 C) 2/27 D) 1/9 E) Can not be determined from the information given.

I have not taken any probability but I figured if the dice was "fair" then there would be 3/6 chance to roll an even number and a 3/6 chance to roll an odd number.

But the probability of rolling an even number is twice that of rolling an odd number

So does this mean the probability of rolling an even number is 4/6 and the probablity of rolling an odd number is 2/6?

I am not sure how to figure out the next part if that is correct though. Can someone help?
• Apr 15th 2010, 11:47 AM
Quote:

Originally Posted by chrisceville22
A six sided die, faces numbered 1 - 6 as usual, is known to be counterfeit. The probability of rolling an even number is twice of the probability of rollong an odd number. What is the probability that, if the die is throw twice, the first roll will be a 5 and the second will be a 6?

A) 2/8! B) 1/18 C) 2/27 D) 1/9 E) Can not be determined from the information given.

I have not taken any probability but I figured if the dice was "fair" then there would be 3/6 chance to roll an even number and a 3/6 chance to roll an odd number.

But the probability of rolling an even number is twice that of rolling an odd number

So does this mean the probability of rolling an even number is 4/6 and the probablity of rolling an odd number is 2/6?

I am not sure how to figure out the next part if that is correct though. Can someone help?

Hi Chris,

I'd imagine answer A) should read $\displaystyle \frac{2}{81}$

Imagine throwing this die 900 times.

600 times we should have an even digit...
300 times we should have an odd digit

hence

$\displaystyle P(E)=\frac{6}{9}$

$\displaystyle P(O)=\frac{3}{9}$

5 is one of the 3 odds, so $\displaystyle P(5)=\frac{1}{9}$

6 is one of the evens, so $\displaystyle P(6)=\frac{2}{9}$

therefore the probability of 5 followed by 6 is $\displaystyle P(5)P(6)$
• Apr 15th 2010, 11:55 AM
chrisceville22
Hmm probably a typo.. Can you please explain with P(5) is 2/9? and why P(6) = 1/9. I am still a bit confused..

Like if 5 is one of the odds.. and the probability for odds is 3/9 then why is it 2/9 to roll a 5? Why not 1/9?
• Apr 15th 2010, 12:04 PM
Quote:

Originally Posted by chrisceville22
Hmm probably a typo.. Can you please explain with P(5) is 2/9? and why P(6) = 1/9. I am still a bit confused..

Like if 5 is one of the odds.. and the probability for odds is 3/9 then why is it 2/9 to roll a 5? Why not 1/9?

sorry Chris,
that was a typo on my part !!
re-edited
• Apr 15th 2010, 12:15 PM
chrisceville22
So it should be to be 1/9 to roll an odd and 1/9 to roll an even?
Then it should be 1/81? which is not an option unless b) was a typo 1/18 should of been 1/81?
• Apr 15th 2010, 12:48 PM
$\displaystyle \frac{1}{9}$ for a particular odd
$\displaystyle \frac{2}{9}$ for a particular even
$\displaystyle \frac{1}{9}\frac{2}{9}=\frac{2}{81}$ for 5,6