View Full Version : help needed...

adamantlyelusive

11-06-2004, 08:47 PM

hi ppl......

would be grateful if u could help me out with these probs...

dont have the certain ans for any of the qs...

1. A cubic object 3"x3"x3" is painted red on all the outside surfaces, including the top and bottom. If the cube is cut into 27 cubes of 1"x1"x1" then

a. how many cubes had 3 sides red painted

b. how many cubes had 2 sides red painted

c. how many cubes had 1 sides red painted

d. how many cubes had 0 sides redpainted

Ans: a.=8, b=12, c.=6, d.=1(not sure)

3. My rack contains 8 Red colour ties, 13 violet colour ties,10 Blue colour ties, 5 Pink colour ties, 4 green colour ties. If there is no electricity and i want atleast two ties of the same colour then how many ties should i take out from my rack?

4. Two trains are leaving from two stations seperated by 50 miles. With a constant speed of 60 miles per hour, the trains are approaching each other on different tracks. if length of each train is 1/6th of a mile, when they meet how much time do they need to pass each other totally?

thanx a ton

bye

calm_J

11-06-2004, 09:47 PM

1- a) and this is can be found if u gave it some little more effort and draw it where the cubes that have 3 sides red are the cubes on the vertices of the cube and the vertices are 8 so a=8

b) the cubes having 2 sides red are the cubes on each edge between the cubes on the vertices found in part (a) noting that if u take 4 cubes on the top there are another 4 on the bottom the there are 2 cubes left and another 2 on right

c) now you have six sides and 1 cube left on each side so the cubes are 6

d) only one cube left in the middle of the larger cube

2- i'm not sure of this but i think that first we have to get the probability of finding one tie of each type in which to know that all ties are 40 .

then

P(r) = 8/40

p(v)=13/40

p(b)=10/40

p(p)=5/40

p(g)=4/40

so if we want to have 2 ties of same type then it's for sure will have a probability less than one of a type then we divide each of those probabilities by 2

P(r) = 4/40

p(v)=13/80

p(b)=5/40

p(p)=5/80

p(g)=2/40

now if we want it to be the probability of 2 among all then we have to multiply all the above probabilities to get the answer...... i hope this is right

3- they will reach each other in the middle of the track however we don't need this info. but the speed is 60/h and the length of train is 1/6 mile then

D=S.T so T = D/S which means that T = 1/360 H = 1/6 min= 10 seconds

thebullfighter

11-06-2004, 09:54 PM

u figured out th 1st one...(ur answers r correct). tip is to draw one face, and multiply wth no. of other possiblities according to th case.. & above tht visualization..

for Q2(3). consider this...

8R, 13V, 10B, 5P, 4G (doesn't matter if there were more of any colour.. u'll see)

start selecting ties One-by-one considering the assumption tht it's one of ur Wrost days.. so take worst case scenerio...

i pick 1-- it's R

1 more-- V (man, why not R ... anyways lets move..)

1more-- B

1more-- P

1more-- G (wow, wat luck... 5till now and no2 of same colour.. god help..)

1more-- YES!!! Yepeeee.. (doesn't matter which colour, another same coloured tie will be there with u already...)

so, total 6 attempts.

Q3.(4) relative speed=60+60=120

total distance to be covered= length of train1+length oftrain2 =2x

hence, time taken= distance/speed=2x/120

x=1/6, so T=10 seconds

calm_J

11-06-2004, 10:07 PM

yeahhh bullfighter u r right. I missed that one. I didn't read the ques. well he was asking about how many ties and i answered about the posibility of ties ..... ok

here it's, let's rephrase the ques. and make it the probability of getting two ties of same color...... does my answer apply?????..... plz feedback

thebullfighter

11-06-2004, 10:40 PM

yeahhh bullfighter u r right. I missed that one. I didn't read the ques. well he was asking about how many ties and i answered about the posibility of ties ..... ok

here it's, let's rephrase the ques. and make it the probability of getting two ties of same color...... does my answer apply?????..... plz feedback

actually for probability u'll have to follow this approach (to be safe)---

Red: 8P2/40P2

V: 13P2/40P2

B: 10P2/40P2

P: 5P2/40P2

G: 4P2/40P2

so, total will be sum of each individual one. but don't calculate... it's not th Q.:)

HTH

calm_J

11-07-2004, 02:39 AM

hi bullfighter,

sorry i didn't get what's meant by 8P2/40P2 and so on.... could u please explain.

thanks in advance.

thebullfighter

11-07-2004, 08:24 AM

hi bullfighter,

sorry i didn't get what's meant by 8P2/40P2 and so on.... could u please explain.

thanks in advance.

8P2 is 8*7 and 40P2 is 40*39

also,

13P4 would be 13*12*11*10

7P3 would be 7*6*5

and similarly...

HTH

calm_J

11-07-2004, 09:46 AM

thanks bullfighter. I got this one but i still can't get how to find the probabilities. please explain in details if possible although i have a math. background but i left these kind of math long time ago so i do have some difficulties with probability, combinations and computations. any help would be appreciated.

thebullfighter

11-07-2004, 03:27 PM

my pleasure.

Probability is defined as total no. of ways of the 'Required' case divided by total number of ways of the same case out of the Total cases available.

So, now if you have 8 ties, u can select two from them in 8*7 ways, I am sure that much is clear with you.

and also, Total possible outcomes of selecting two ties out of all the given ties will similarly be 40*39. (ie, for first tie, there are 40options, and for second 39options or ways.)

so if required is 2 red ties... 2 red ties can be selected from the 8 given in 8*7 ways, and 2ties selected from total 40ties be done in 40*39 ways.

so, probability of 2red ties from 40 ties is reuired/total = 8*7/40*39

also, in probability if 'OR' is there we 'Add' n if 'And' is there we 'Multiply' individual probabilities.

So, as here it's 2Red ties or 2Violet ties or...... so on..

so, we add.

Consider, required 1red tie And 1Violet tie. Solve it for me.:)