probability

[edt966]

I understand pairwise mutually exlusive events. When I read this problem I first thought the answer was 1/2 of getting an even number 2, 4, or 6 on a roll. But the correct answer is 4/7. Can anyone suggest a similarly worded problem or reword this one? The question was:

"A die is loaded in such a way that the probability of the face with j dots turning up is proportional to j or j = 1,2,3,4,5,6. What is the probability, in one roll of the die, that an even number will turn up?" Answer 4/7.

Does the word "loaded" in this question generally mean "a purposefully altered die?" You know, like in the back alley scenes of a movie..."loaded dice."

Also, what is meant by "...j dots turning up is 'proportional'..."?

Thanks, Ed

 

[royhaas]

The probability "proportional" just means that the probability of throwing a six is six times the probability of throwing a one. Since the probabilities must add to 1, and 1+2+3+4+5+6=21, you must have that p(i)=i/21, for i=1,...,6. From this it's easy to calculate the probability of an even number.

 

[soroban]

Hello, Ed!

"Loaded dice" is simiarly to a "stacked" deck.

They have been altered in advance (secretly)
\(\displaystyle \;\;\)so that the outcomes do not have the usual probabilities.

Here's my trademark: baby-talk.


Rolling an "honest" die, the outcomes {1, 2, 3, 4, 5, 6} are equally likely.
\(\displaystyle \;\;\;P(1)\,=\,P(2)\,=\,P(3)\,=\,P(4)\,=\,P(5)\,=\,P(6)\,=\,\frac{1}{6}\)

The problem is identical to:
\(\displaystyle \;\;\)There are six cards in a bowl. They are numbered "1" to '6".
\(\displaystyle \;\;\)You draw one card at random.

In both scenarios, \(\displaystyle P(\text{even})\:=\:\frac{3}{6}\:=\:\frac{1}{2}\)


Suppose the bowl contains: {1, 2,2, 3,3,3, 4,4,4,4, 5,5,5,5,5, 6,6,6,6,6,6}
\(\displaystyle \;\;\)The bowl has 21 cards: 9 odd numbers and 12 even number.

Hence: \(\displaystyle \,P(\text{even})\:=\;\frac{12}{21}\:=\:\frac{4}{7}\)


We can see that: \(\displaystyle \,P(2)\,=\,\frac{2}{21}\,\) and \(\displaystyle \,P(1)\,=\,\frac{1}{21}\)
\(\displaystyle \;\;\)That is, a "2" is twice as likely as a "1".

Similarly, "3" is three times as likely as "1",
. . . . . . . ."4" is four times as likely as "1", and so on.

And that is what was meant by "the probabilities are proportional".