## Wednesday, June 3, 2009

### The Egyptians have Scribes.

Hi guys! This is Pacifico, the scribe for today!

Today, we delved deeper into the cancer question, and other things; but mainly the cancer question. We also learned a new formula! And we also get to know another way to solve a certain probability question. Things to remember first:

v Don’t forget to accept the invite to the DEV blog!

v GET ENOUGH SLEEP FOR THE EXAM!

v STRESS IMPEDES LEARNING!

v In A|B, | is a shortcut for the phrase, “…given that we already know…”

So, let’s get this scribe started!

For this part, I will only talk more about Q. #4 because Q. #1-3 is a no-brainer for us. Q. #4 starts the semi-complicated things, notably c).

In this slide, it tells us that the probability that person who tests positive for cancer has cancer. Mr. K. used the formula P(C|P) = (4900 ÷ 2400) = 19.76%

P(C|P) is said as: probability of people that have cancer given that they’re already positive.

How did the heck did he get that, you ask?

C = has cancer

H = doesn’t have cancer

P = tested positive

N = tested negative

P(C|P) = P(CP) ÷ [P(CP) + P(HP)] or

Probability of people that have cancer given that they’re already positive is equal to the probability of people that have cancer that is tested positive divided by the sum of the probability of the people that don’t have cancer but is tested positive.

Therefore,

P(C|P) = ( 0.005 × 0.98) ÷ [(0.005 × 0.98) + (0.995 × 0.02)] = 0.1976

So the answer to the question: What is the probability that a person who tests for positive for cancer has cancer is 19.76%

This slide is asking for P(H|N). By reading the question, we know that it’s the same as the last question. So we apply the knowledge that we know already and solve the question!

P(H|N) = P(HN) ÷ [P(HN) + P(CN)]

The next slide is also the same question as the last 2, so I’m pretty sure that you can figure this out on your own (unless this scribe is sucky—which by the way you have to tell me so I can change it!)

Ah… This question…

To find the probability that the couple has at least 2 girls assuming that having a boy or a girl is equally likely events…

We used permutations, from the unit Combinatorics, to find the number of ways of getting 2 girls.

You could use a tree diagram to find the answer, like what is shown below, but that’ll take a long time and it’s going to waste space.

Therefore,

GGBB à 4 “slots” (number of letters) and 2 non-distinguishable object that both have 2 same things.

number of ways to get 2 girls = 4! ÷ (2!2!) = 6

And then, we figure out the probability of getting 2 girls and 2 boys.

Therefore,

P(GGBB) = (½)(½)(½)(½) = 1/16

After that, we multiply the P(GGBB) and the number of ways to get 2 girls.

Therefore,

P(2G) = 6 × (1/16) = 6/16 = 3/8

To compress all that: P(2G) = [4! ÷ (2!2!)] × [(½)(½)(½)(½)] = 3/8

The original question was, “find the probability that the couple has AT LEAST 2 girls.” Therefore, we also calculate the P(3G) and P(4G), which is shown in this slide.

Then to get the final answer, we add P(2G), P(3G) and P(4G) together.

HOLD ON!

There’s another easier way to find the answer!

Use the COMPLIMENT PROBABILITY

Therefore,

P(at least 2 girls) = 1 – [P(1G) + P(0G)]

= 1 – [(1/2)^4 × (4+1)]

= 11/16 <-- answer is the same as adding P(2G), P(3G) and P(4G)

together

That concludes my scribe post…

Next scribe is… REBECCA! Hurr durr…

Postedit: Sorry for the lateness! I was doing it in Dreamweaver then I fell asleep halfway of the scribe post!