No math yet?? PCAT??

[firstgeneration]

If i haven't taken calculus should i attempt to take the PCAT test or not?? because it is 22% of the test??? any suggestions

---

Members don't see this ad.

 

[xshooshix]

If i haven't taken calculus should i attempt to take the PCAT test or not?? because it is 22% of the test??? any suggestions

You should check out those self-help books. I haven't looked at a Calculus book in a year and a half, so I DEFINITELY need some reviewing myself. Do a search in the PCAT forum and type in 'calculus'. You'd be surprised with the results.

 

[Farmercyst]

Moved to PCAT discussion thread.

 

[Magikarp]

I think you should still take it. From what I remember from the test, the calc was very very basic, like first derivaties and integrals, stuff you can learn in one study session.

As for trig/geometry...

 

[firstgeneration]

I think you should still take it. From what I remember from the test, the calc was very very basic, like first derivaties and integrals, stuff you can learn in one study session.

As for trig/geometry...

thanks for the suggestions i would take in consideration

 

[firstgeneration]

You should check out those self-help books. I haven't looked at a Calculus book in a year and a half, so I DEFINITELY need some reviewing myself. Do a search in the PCAT forum and type in 'calculus'. You'd be surprised with the results.

Thank you for the input. I actually so those discussions and tips that were given on the forum previously. I was actually nervous about others things that can be on the test besides the tips.

 

[Knickerbocker]

I think you should still take it. From what I remember from the test, the calc was very very basic, like first derivaties and integrals, stuff you can learn in one study session.

As for trig/geometry...

The official page says that the new version of the test no longer has any geometry problems.

 

[Chokita]

Help me to solve this problem:
1. Determine the probability of having 1 girl and 3 boys in a 4-child family. (Answer 1/4)


firstgeneration, in case if you want to feel comfortable with differentiation and integration before the test, you might need to read this: http://www.mhhe.com/math/calc/smithminton2e/cd/

 

[omnione]

Help me to solve this problem:
1. Determine the probability of having 1 girl and 3 boys in a 4-child family. (Answer 1/4)

To go through all the possibilities that can happen assuming each child comes through a separate birth:

B-B-B-B= 1
G-B-B-B = 4
B-G-G-G = 4
B-B-G-G = 6
G-G-G-G = 1

4/16 = 1/4

To be honest, I would've gotten this problem wrong on the PCAT because my knee-jerk response would have been 1/16 [=(1/2)^4] but I neglected the fact that the girl could have been born first, second, third, or fourth by assuming that the girl could only have been born first.

EDIT: So, the best way to approach a probability problem is to manually or use some formula to consider all the possibilities driven by logic first. So, ask yourself "what can happen?" when doing a probability problem.

 

[xshooshix]

To go through all the possibilities that can happen assuming each child comes through a separate birth:

B-B-B-B= 1
G-B-B-B = 4
B-G-G-G = 4
B-B-G-G = 6
G-G-G-G = 1

4/16 = 1/4

Can you explain how you got 1, 4, 4, 6, 1?

 

[Wow Strong Armz]

Can you explain how you got 1, 4, 4, 6, 1?

easier way i think this is correct is to use the combination formula

so 4!(#of people in family)/ (3!(since there are 3 boys that are indistinguishable)) and you will get 4 diffrent combinations then u can ask ur self what is the prob of getting a girl which would be 1/4. dont quote me on this until somone says my logic is right but thats how i would do it

 

[omnione]

Can you explain how you got 1, 4, 4, 6, 1?

Sure thing....basically, I just compiled all the possibilities for each outcome that is possible:

To start:

B = Boy
G = Girl

Outcome 1: All Boys
B-B-B-B
-There's only one way to get all four kids to be boys

Outcome 2: One girl, three boys
-Now, there are four ways this can happen. Basically, it includes all the combinations where a girl is born in the first, second, third, or fourth. I should've made this more clear in my first post:

G-B-B-B
B-G-B-B
B-B-G-B
B-B-B-G
-I would have gotten this problem wrong on the real PCAT because I forgot to account for the fact that the time that the girl is born doesn't matter

Outcome 3: Two Boys -Two Girls
-There are six ways this can happen:
B-B-G-G
G-G-B-B
B-G-G-B
G-B-B-G
G-B-G-B
B-G-B-G

Outcome 4: One boy - three girls
-See Outcome 2 but switch around boys and girls and you get four possibilities

Outcome 5: Four girls
-See Outcome 1 but consider the lone possibility of getting four girls

So, there sixteen total ways parents can have four kids. Of those sixteen, only four fit the requirement that one is a girl and three are boys. 4/16 reduces to 1/4.

 

[omnione]

easier way i think this is correct is to use the combination formula

so 4!(#of people in family)/ (3!(since there are 3 boys that are indistinguishable)) and you will get 4 diffrent combinations then u can ask ur self what is the prob of getting a girl which would be 1/4. dont quote me on this until somone says my logic is right but thats how i would do it

Almost. Combinations are actually n!/[(n-k)! * k!] and permutations are n!/(n-k)!

Your results actually tells you how many ways three boys can be born out of four children. So, that also tells you how many ways one girl may be born. Still, you have to come up with the total number of ways the parents can have four kids.

 

[Wow Strong Armz]

Almost. Combinations are actually n!/[(n-k)! * k!] and permutations are n!/(n-k)!

Your results actually tells you how many ways three boys can be born out of four children. So, that also tells you how many ways one girl may be born. Still, you have to come up with the total number of ways the parents can have four kids.

i c, if i remb correctly almost all the prob questions had to do with deck of cards. know ur queens and kings