Thread: probability

A dice is loaded so that the numbers 1, 2, and 3 are equally likely to appear. 4, 5, and 6 are also equally likely to appear, but 1 is three times as likely as 4 to appear. What is the probability of getting an odd number?

Having trouble with this one.

Cheq

Originally Posted by cheqtel

A dice is loaded so that the numbers 1, 2, and 3 are equally likely to appear. 4, 5, and 6 are also equally likely to appear, but 1 is three times as likely as 4 to appear. What is the probability of getting an odd number? Having trouble with this one.

I am not surprised!
There is no such a thing as a dice. There is a die.
There are two dice.

Ok. Still if it is a die. and I agree that it is not a dice. What would the answer be?

pwr

Originally Posted by cheqtel

A dice is loaded so that the numbers 1, 2, and 3 are equally likely to appear. 4, 5, and 6 are also equally likely to appear, but 1 is three times as likely as 4 to appear. What is the probability of getting an odd number?

Having trouble with this one.

Cheq

As the sum of the probabilities must be 1, let the prob of a 4 showing be p,

then:

(3*p + 3*p + 3*p) + (p + p + p) = 12p =1.

and I hope you can solve this from here.

RonL