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SLB1966 · May 22, 2011, 04:31 PM

I'm really stumped on this one:

"The probability of C is 0.4. The conditionaly probability that C occurs given that D occurs is 0.5. The conditional probability that C occurs given that D does not occur is 0.25.

A. What is the probability that D occurs?
B. What is the conditional probability that D occurs given that C occurs?

I honestly don't know where to start with this one. Anyone got a clue?

jcaron2 · May 22, 2011, 08:41 PM

You can write out the conditional probability equation for events C and D:

$P(C|D)=\frac{P(D \cap C)}{P(D)}=0.5$

You can also write the equation for C and (not D):

$P(C|\overline{D})=\frac{P(\overline{D} \cap C)}{P(\overline{D})}=\frac{P(\overline{D} \cap C)}{1-P(D)}=\frac{P(C)-P({D} \cap C)}{1-P(D)}=\frac{0.4-P({D} \cap C)}{1-P(D)}=0.25$

This gives you two equations with two unknowns, $P(D)$ and $P(D \cap C)$ . You can use the two equations to solve for $P(D)$ (which answers part A). Then solve for $P(D \cap C)$ (which is the same as $P(C \cap D)$ of course), which you can then use, along with $P(C)$ , to compute $P(D|C)$ to answer part B.

Make sense? If you're confused about the step in the second equation where I use $P(\overline{D} \cap C)=P(C)-P({D} \cap C)$ , just draw out a Venn diagram and you'll see that it's true.

Unknown008 · May 23, 2011, 10:25 AM

And I prefer probability trees on question like these so that there are no simultaneous equations. ;)

Use the first probability tree for the first part. You know that the probability that C occurs is either going through D then C, or through $\bar{D}$ then C. Then, let the probability that D occurs (in the first upper branch) is d. The other lower branch, $\bar{D}$ will be 1-d.

Hence you get:

$(d\ \times \ 0.5) + ((1-d)\ \times \ 0.25) = 0.4$

or if you prefer set notation...

$$P(D) * P(C|D)$ + $P(\bar{D}) * P(C|\bar{D})$ = P(C)$

Which if you work it out on a Venn Diagram works out good :)

Once you get D, you can get everything easily. For the second part, you just need the upper branch. You know the result of going through this path, using the previous values of P(C) and P(D) and the definition of the probability of D occurring given C occurred. :)

SLB1966 · May 23, 2011, 04:32 PM

To be honest, even with the help you've given me, I still don't see it. And, my instructor showed me another route... still doesn't click. And, I did draw a Venn diagram... didn't help. Guess statistics isn't for me!

SLB1966 · May 23, 2011, 04:32 PM

To be honest, even with the help you've given me, I still don't see it. And, my instructor showed me another route... still doesn't click. And, I did draw a Venn diagram... didn't help. Guess statistics isn't for me!

SLB1966 · May 23, 2011, 04:33 PM

Also, thanks for trying. Should've done this right after high school, not 27 yrs later!

SLB1966 · May 23, 2011, 04:33 PM

Also, thanks for trying. Should've done this right after high school, not 27 yrs later!

jcaron2 · May 23, 2011, 05:25 PM

If it's any solace, that's NOT an easy statistics problem. There are lots and lots of people (including more "traditional" college students) who would have trouble with it.

Anyway, let's keep going using my method based on the two equations I gave:

$\frac{P(D \cap C)}{P(D)}=0.5$

$\frac{0.4-P({D} \cap C)}{1-P(D)}=0.25$

First, let's rearrange the top one:

$P(D \cap C)=(0.5)P(D)$

Now we can substitute the value on the right in place of the value on the left in the second equation:

$\frac{0.4-(0.5)P(D)}{1-P(D)}=0.25$

Now let's just solve it for P(D):

$0.4-(0.5)P(D)=(0.25)(1-P(D))$

$0.4-(0.5)P(D)=0.25-(0.25)P(D))$

$0.4-0.25=(0.5)P(D)-(0.25)P(D))$

$0.15 = (0.25)P(D)$

$0.6 = P(D)$

There's the answer to part A.

Now we can substitute that value back into the first equation to get the other value we're looking for:

$P(D \cap C)=(0.5)P(D)=(0.5)(0.6)=0.3$

Now we can finally find the answer to part B:

$P(D|C)=\frac{P(C \cap D)}{P(C)}=\frac{0.3}{0.4}=0.75$

And just to make sure the answers are right, let's substitute everything back in to find the probability of C given D, and make sure it comes out to 0.5 like the problem initially stated:

$\frac{P(D \cap C)}{P(D)}=\frac{0.3}{0.6}=0.5$

Yup! It worked out. The answers must be right.

Does that make any more sense, or are you still lost in one of the initial steps? I'll be happy to explain in more detail if I skipped too much.

Unknown008 · May 24, 2011, 07:57 AM

Um... part A, 0.5P(D) - 0.25P(D) = 0.25P(D)

:)