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No math yet?? PCAT??
Started by firstgeneration
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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.
Moved to PCAT discussion thread.
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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
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.
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.
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/
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/
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.
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
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.
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
