Calculating Probability of Even Numbers with Loaded Die

[Bucs44]

Homework Statement

A die is loaded so that the numbers 2 through 6 are equally likely to appear but that 1 is 3 times as likely as any other number to appear. What is the probability of getting an even number?

Homework Equations

This is where I become lost - I'm not sure how to get the outcome.

The Attempt at a Solution

P(2) + P(4) + P(6) = 1/8 + 1/8 + 1/8 = I think the answer is 3/8 but can't tell you how I got that. I don't understand the P(1) + P(2) + P(3) thing. Could someone help me to understand this?

 

[matt grime]

What P(1)+P(2)+P(3) thing?

You know that there are 6 outcomes

P(2)=P(3)=..=P(6), P(1)=3P(2), and P(1)+P(2)+..P(6)=1.

So from this you can find P(2), and hence P(2 or 4 or 6)=P(2)+P(4)+P(6).

 

[Bucs44]

What P(1)+P(2)+P(3) thing?

You know that there are 6 outcomes

P(2)=P(3)=..=P(6), P(1)=3P(2), and P(1)+P(2)+..P(6)=1.

So from this you can find P(2), and hence P(2 or 4 or 6)=P(2)+P(4)+P(6).

That's just it - I don't understand this. I guess I see where P(2)*P(3) = P(6) but where I get lost is P(1)=3P(2), and P(1)+P(2)+..P(6)=1.

 

[jermanie]

Load a die and die cast yourself.

 

[Bucs44]

Load a die and die cast yourself.

Geez - thanks for the great input there! Some of us aren't math wizz's.

 

[HallsofIvy]

Load a die and die cast yourself.

The montecarlo method!

That's just it - I don't understand this. I guess I see where P(2)*P(3) = P(6) but where I get lost is P(1)=3P(2), and P(1)+P(2)+..P(6)=1.

You "see" the one part I don't see, and are lost on the easy ones!

The problem says "1 is 3 times as likely as any other number to appear". If the probability a 2 will occur is P(2) and 1 is "3 times more likely", then P(1)= 3P(1). And, of course, since some number must occur the probability some number must occur is 1. That is, the sum of all the individual probabilities is 1. That's generally true for any probability distribution- the sum of probabilities of all "elementary" events is 1.

If you let p be the common probability of 2- 6, P(2)= P(3)= P(4)= P5)= P6) and P1)= 3p so you must have 3p+ p+ p+ p+ p+ p= 1. Solve that for p and the rest is easy.

But what in the world makes you think that P(2)*P(3)= P(6)? They are not, P(2)= P(3)= P(6). Are you thinking that 2*3= 6? That has nothing to do with probabilities.

 

[Bucs44]

So P(2) + P(4) + P(6) = 1/8 + 1/8 + 1/8?