Hello,
I am trying to figure out if I did my homework correctly.

The question is:
Consider the experiment of rolling two dice. The outcome on each die can be any number of dots from 1 to 6. The dice are specially manufactured such that the probability of an even number (of dots) is twice that of an odd number. Let x= the sum of the number of dots on the two dice.

Determine P(X<4)

a. 0.0833
b. . 3330
c. . 0617
d. .1111

I chose d, and I obtained this from (4/36)(1/3)+(4/36)(1/3). Is this correct?


Additionally, I had another question that I need help with.

Three events X,Y,and Z represent a partition of the sample space of a random experiment. A is an event defined on the sample space. The following probabilities are given:

p(x)= .2
p(y)=.35
p(z)= .45
(Probability of A, Given X)= .15
Probability of Y AND A = .07
Probability of A, given Z= .4

I do not believe this is enough data to complete the problem. Please tell me if I am wrong.

Thanks!

1. Nope. For 1. I get C, 0.0617.

This is how I do it, the simple way:

Probability of having P(X<4) = P(X=2) + P(X=3) right?

How can you get P(X=2)? It's when you have (1, 1)

What is the probability of having 1 from the dice?

Let's say the probability is x for odd numbers.
It's 2x for even numbers.
Total is 3x.
So, x = 1/3.

1/3 is the probability of having an odd number. That of having a 1 therefore is 1/9, that for having a 3 is 1/9 and that for having a 5 is 1/9.
Same for even. P(2) = 2/9, P(4)=2/9, P(6)=2/9.

Coming back to the question,

P(X<4) = P(1, 1) + P(1, 2) + P(2, 1) = \(\frac19\)\(\frac19\) + \(\frac19\)\(\frac29\) + \(\frac29\)\(\frac19\)

2. For this one, what is the question? Do you need to find the P(A)?

P(X) = 0.2

P(Y) = 0.35

P(Z) = 0.45

P(A|X) = 0.15 = \frac{P(A \cap X)}{P(X)}

P(Y\cap A) = 0.07 = P(Y) \times P(A|Y)

P(A|Z)= 0.4 = \frac{P(A \cap Z)}{P(Z)}

If you draw a probability tree, you can easily see how to get P(A).

Let's say, X, Y and Z are urns with a ticket. A is the winning ticket. P(A) is the probability you get a winning ticket, and that can be either from urns X, Y or Z.

What is the probability to get a winning ticket from X? It's P(A \cap X). A winning ticket from Y and Z are P(A \cap Y) and P(A \cap Z) respectively. So add the three of them, to get P(A).

One way to make things easier on stuff like #2 when you have more than 2 things is to make a joint probability table, which is basically taking one set of variables across the top of a chart and the other set down the sides.

Notice out to the right is the probability of X, Y & Z, which total up to 1.00. (Which they should.) We'd also put the A, B & C across the bottom, which would total to 1.00 as well, but we don't know those numbers. Nor do we know anything about B & C at all, but you could use this chart to figure that out as well. (There's no rule saying something has to be every one of them, as long as they add up to the bottom right as 1.00.)

The yellow spaces are all intersections. Hence, the intersection of Y and A is .07, therefore it goes into that space where they cross. You do not need an equation to solve that as it's given.

Then use the other two "given" equations and a little algebra to solve for the intersections for A & X, and A & Z, and just plug those into where they cross. You can easily see from that what the total of A would be. If you had similar problems, you could use this as well, to solve for different intersections, or the total of C, or even a union (using your addition equation).