math problem

Started by Doctor PJ
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Doctor PJ

Doctor PJ

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can someone show me how this is done. With one die, what is the probablity of throwing two fours in five attempts?

Answer is: 10 * 125/ 36 * 216
 
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klutzy1987 said:
Crap i thought it was a coin and heads tails.

It would be 5c2 * (1/6)^2 * (5/6)^3

10* (1/36) * (125/216)

1250/7776---->625/3888

Those are stupid numbers.
Click to expand...


Thanks bro! I actually understand this problem as you did some similar problems. You freaked me out for a second there 😀.
 
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Thundercatz said:
Thanks bro! I actually understand this problem as you did some similar problems. You freaked me out for a second there 😀.
Click to expand...

LOL sorry i gotta be more careful reading the questions. But then again for me it makes no diff anymore as i already took my DAT lol.
 
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klutzy1987 said:
Crap i thought it was a coin and heads tails.

It would be 5c2 * (1/6)^2 * (5/6)^3

10* (1/36) * (125/216)

1250/7776---->625/3888

Those are stupid numbers.
Click to expand...

Ive never heard of the success failure formula.. so..
I gather its 5c2 * chance of success^(how many times) * chance of failure^(how many times)

can som1 explain to me what 5c2 is?
 
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I have a question about this..

in achiever, the question is What will be the likelihood of having exactly 1 boy in a family planning for three children?

what i did was that 3C1*(1/2)*(1/2)^2

but actual answer is 3C2*(1/2)*(1/2)^2

why is that???

help please..
 
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jjw822 said:
I have a question about this..

in achiever, the question is What will be the likelihood of having exactly 1 boy in a family planning for three children?

what i did was that 3C1*(1/2)*(1/2)^2

but actual answer is 3C2*(1/2)*(1/2)^2

why is that???

help please..
Click to expand...

I am a little confused by your notation of 3c?. But I can explain it how I know it.


You want AT LEAST 1 boy in 3. So that means it can be

Boy, Girl, Girl (1/2) ^3

Girl, Boy, Girl (1/2)^3

Girl, Girl, Boy (1/2) ^3


Those are the separate probabilities of each event occuring... so you have...

1/8 chance of one of those happening.
To get the total probability

since you can ahve either 1.. OR 2... OR 3...

you add..
you get 3/8 as your final answer.

did that help?
 
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priyerz said:
I am a little confused by your notation of 3c?. But I can explain it how I know it.


You want AT LEAST 1 boy in 3. So that means it can be

Boy, Girl, Girl (1/2) ^3

Girl, Boy, Girl (1/2)^3

Girl, Girl, Boy (1/2) ^3


Those are the separate probabilities of each event occuring... so you have...

1/8 chance of one of those happening.
To get the total probability

since you can ahve either 1.. OR 2... OR 3...

you add..
you get 3/8 as your final answer.

did that help?
Click to expand...

what i was referring was nCr rule... and i dont understand what you meant by you can have either 1.. or 2 or 3




and one more question on achiever
If 1/3 of the cars driven by residents in a specific town are Japanese made, what is the probability of seeing at least one Japanese car out of every three automobiles on the road?

and this is the explaination.. which i don't understand..



Correct Answer: B
The probability of NOT seeing any Japanese car out of every three automobiles is (2/3)3 = 8/27. {Multiplication Rule}
Using complementary rule, the probability of seeing at least one Japanese car out of every three automobiles is therefore,
1 – (8/27) = 19/27
 
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all three of those are possibilities... so you can have either one.

first part

you can ahve

1 boy AND 1 girl AND 1 girl as an option.. you multiply the probabilities...


since you can only have.... one of the options of order of kids...
its either choice 1.... OR choice 2 (meaning what order of kids)... so since it is EITHER OR, you add the probabilities....

its jsut important to remember to multiply it by 3 becuase there are 3 possible WAYS to get that same outcome.

unless they say, have a boy firends adn then two girls... then its 1/8...
but if its any of those combinations, your probability increases... because you can have them in more different ways...

confusing? i may have explained terribly.
 
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haha jokes

guess achiever just takes questions from random workbooks word by word.

I solved a question exactly worded (except it was 2 boys not 1) last night from this booklet someone gave me.
 
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jjw822 said:
what i was referring was nCr rule... and i dont understand what you meant by you can have either 1.. or 2 or 3

and one more question on achiever
If 1/3 of the cars driven by residents in a specific town are Japanese made, what is the probability of seeing at least one Japanese car out of every three automobiles on the road?

and this is the explaination.. which i don't understand..



Correct Answer: B
The probability of NOT seeing any Japanese car out of every three automobiles is (2/3)3 = 8/27. {Multiplication Rule}
Using complementary rule, the probability of seeing at least one Japanese car out of every three automobiles is therefore,
1 – (8/27) = 19/27
Click to expand...

Same thing here.. you need optoins---

since AT LEAST 1 is japanese... you can have these options
you see 1 J, 2J, or 3 J

J= japanese = 1/3
O = other = 2/3

J , J, O = (1/3) (1/3) (2/3) = 2/27

O J J = 2/27

J O O = 4/27

O J J = 4/27

O J O = 4/27

J J J = 1/27

JOJ = 2/27


add those you get
19/27

tell me if i didn't explain it well
 
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Sea of ASH said:
yea what is 5c2????
Click to expand...


the nCk notation refers to:

n!/(n-k)! so 5C2 would be 5!/(5-2)! which basically equals:

(5x4x3x2x1)/(3x2x1)

what i don't understand is how do you know when it is a permutation or a combination? as in how can you tell whether order matters or not so that you divide by k! as well..?
 
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jjw822 said:
what i was referring was nCr rule... and i dont understand what you meant by

hoyas19 said:
the nCk notation refers to:

n!/(n-k)! so 5C2 would be 5!/(5-2)! which basically equals:

(5x4x3x2x1)/(3x2x1)

what i don't understand is how do you know when it is a permutation or a combination? as in how can you tell whether order matters or not so that you divide by k! as well..?
Click to expand...


well i dont think you can use it for this problem.... but
order matters if they say it matters....
like if htey say if they want to have a boy and then two girls....
or they want jenny to sit in teh second seat....


You just have to read VERY carefully.... (i ahvnt taken dat so i dont know if these show up or not... anyone have input?) but basically you need to decide

is this probability??
is this combining probabilities? aka when order DOES NOT matter...

is this combining probabilities of different things?
have 3 browns fans and 4 steelers fans, probability of getting 2 steelers and 2 browns fans?....
here order DOES NOT matter! so you have to use both! also... in a problem like this.. itis without putting those people back...

example...
if you see 1/3 japanese cars--- it is always 1/3

BUT IN MY EXAMPLE:

you have say the option of

B B S S

so you get

(3/7) (2/6) (4/5) ( 3/4)


You have to make sure you do not add that person back, becuase you have already picked them!!

Doing practice problems is really the only way to get it... but I hope i clarified some main points that you can't mess up.. or you'll def et the wrong answer.


also if it says AT LEAST, or NO MORE THAN.... make sure you do the options with more or less, respectively.

fliping coin and having kids problems are usually exactly the same = always 1/2 chances...


Here is an example see if you can figure it out...

It's sortof a trick for a heads up.. i'll put up the answer after... but its a good problem to see if you get all of those.

A bin has 10 balls. 2 red 2 blue 2 yellow 2 orange 2 green. What is the probability, if 2 balls are randomly selected one at a time, without being put back, that both balls will be the same color?
Click to expand...
 
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also, that equation works for how many possibilities there are.
Someone correct me if i am wrong, but I don't think its the same thing as probability.

that equation refers to a problem like...

there are 4 guys. we want to pick 2 of them.
how many options do we have to pick any 2 of them?

4! / 2! (4-2!)
we have 6 options

prove it:
A B C D

AB
AC
AD
BC
BD
CD

this is different than saying

what is the number of ways we can arrange these boys?
4*3*2*1 = 24
not going to write them out... but you get it..

also different than saying

what is the chance that out of 2 guys with blue hair and 2 with brown
we will pick 1 with brown and 1 with blue

options =

Blue, Brown (1/2) (1/2)
Brown Blue

Probability = 2/4 = 1/2

THIS is ALSO different han saying, probability of picking a guy with blue hair and then one with rown

one option of
blue, brown = 1/4




permutation is when order matters
combination is when it doesnt
remember it this way

permutation u use that formula

combination.. you ahve diff options, so like in my previous examples...

you ned to ADD the probabilities of the diff examples...
makes sense?

typed it quick sorry bout typos
 
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