Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The normal probability of a number in a regular die (6 faces) is $\dfrac{1}{6}$. Let in an addicted [that is, "loaded"] die, the probability of a even number (2, 4 and 6) be twice the normal probability;

I've got such outcome: $regular\space probability \space on \space evens \space is \space \dfrac{3}{6} $, doubling it, it would become $\dfrac{6}{6}$, in other words, a certain event, it sounds strange to me, is that right ?

Thanks in advance;

share|improve this question
1  
ACCEPT SOME ANSWERS! –  Gerry Myerson Apr 12 '12 at 0:45
1  
I've just figured it out now, I'm starting doing it. –  aajjbb Apr 12 '12 at 0:48
3  
If the probability of rolling an even number on the loaded die really is twice what it would be on a fair die, then yes: with the loaded die you are certain to roll an even number. –  Brian M. Scott Apr 12 '12 at 0:53
2  
I'm not sure if others use "right" in the context you imply when you say "right event". You should say "certain event". –  David Mitra Apr 12 '12 at 0:57
    
yes, I meant "certain". Thanks –  aajjbb Apr 12 '12 at 1:52
show 2 more comments

1 Answer

up vote 2 down vote accepted

You might check the wording of the question. If the probability of each even number is twice that of a normal die, you are correct. If the probability of each even number is twice that of each odd number, the result is different-then the evens come up $2/9$ each for a total probability of $6/9=2/3$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.