Proability (1 Viewer)

[MyLuv]

I have some question in Proability but i'm not sure about the answer ,can u guys check it for me??? Thx.This is the 1st one:

Research has shown that 90% of 4 Unit students, 70% of 3 Unit students and 40% of 2 Unit students pass a particluar first year course at University. If the students studying this course are evenly distributed from each of the levels of Mathematics, find the probability that two students chosen at random will pass this course.

This is my answer:
Proab. for 1 student to pass= 1/3*0.9+1/3*0.7+1/3*0.4=2/3
---> 2 student pass= 4/9
Can U guys check it,more's coming

 

[MyLuv]

2nd one

Fred has 3 uniform tetrahedra(triangular pyramid).Each of them has 1 face black,1 green,1 white,1 red.When tossed on to a table,3 faces can be seen.If the proability of any colour not being seen is equally likely,what is the proability that:
i) no black face can be seen?
ii) exactly 2 black faces can be seen?
iii) at least 2 red faces can be seen?
iv) 3 white and only 1 green face can be seen?

My answer:
i) (1/4)^3
ii)3*(3/4)^2*1/4=27/64
iii)3*(3/4)^2*1/4 + (3/4)^3=27/32
iv) (3/4)^3*[3*(1/4)^2*3/4]=243/4096
plz help

 

[MyLuv]

3rd one

If a series of 5 games played by 2 equally matched teams,A and B,the team wins 3 first game is champion
i) If B win the 1st two,what is the proability that team A is the champion?
ii) If A has won 1 game,what is the proability team A is the champion?

i) 1/2^3=1/8
ii) 1/2^2+1/2^3+1/2^4=7/16

 

[inasero]

2nd one


quote:
--------------------------------------------------------------------------------
Fred has 3 uniform tetrahedra(triangular pyramid).Each of them has 1 face black,1 green,1 white,1 red.When tossed on to a table,3 faces can be seen.If the proability of any colour not being seen is equally likely,what is the proability that:
i) no black face can be seen?
ii) exactly 2 black faces can be seen?
iii) at least 2 red faces can be seen?
iv) 3 white and only 1 green face can be seen?
--------------------------------------------------------------------------------


My answer:
i) (1/4)^3
ii)3*(3/4)^2*1/4=27/64
iii)3*(3/4)^2*1/4 + (3/4)^3=27/32
iv) (3/4)^3*[3*(1/4)^2*3/4]=243/4096
plz help

yep the frist three are right...but for the fourth one consider it this way....regardless of which faces can be seen, onle one green face can be seen...so that means in the other two pyramids then the green MUST lie on the bottom, no question about it....
which leaves the third pyramid to consider...
we know white is visible aNd green is ALSO visible (cos it cant lie on the bottom), leaving two possible faces....so therefore u obtain
P=(1/4)(1/4)(2/4)(3)
=3/32

we multiply by three to account for the pyramids that couldve been the odd one out...

 

[MyLuv]

Ok,I see...coz in this case the proablitity for each colour is not idependent...silly me

 

[underthesun]

Originally posted by MyLuv
3rd one

i) 1/2^3=1/8
ii) 1/2^2+1/2^3+1/2^4=7/16

I got :

(i) (21/56) and (ii) (70/126)

 

[MyLuv]

Originally posted by underthesun
I got :

(i) (21/56) and (ii) (70/126)

Can u tell me how to get it???

 

[Constip8edSkunk]

for the 3rd 1 i got

i) (1/2)^3=1/8

ii) (1/2)^2+ 2*(1/2)^3 + 3*(1/2)^4=11/16

 

[MyLuv]

Originally posted by Constip8edSkunk
for the 3rd 1 i got

i) (1/2)^3=1/8

ii) (1/2)^2+ 2*(1/2)^3 + 3*(1/2)^4=11/16

ye right, I got the same thing(after redo it )

 

[underthesun]

How about this:

B won the first two:

there can be 8 more games left: which is

AAAAABBB

where A is A wins, B is B wins. Now, if we put the last slot of the letter as B, then A will win, because it will reach 5 As first. Hence, the chance for A to win is

7!/(5!*2!) <-- the total arrangements with the last letter being B
8!/(5!*3!) <-- total outcomes.

At lest that's how I wor this thing, but your method looks as valid as well..

Far out.. what's wrong with boredofstudies? everything seems to slow down on my browser wheneverI visit this place.. I type faster tthan what the textbox displays,. making it hard to see what I am typing..

I think I got infected by a virus or my browser hicjacked...
Bleh!

 

[MyLuv]

Originally posted by underthesun

there can be 8 more games left: which is
AAAAABBB

Why 8???
Anyway,here is another 1:

i) in how may way can 10 student group into 2 team of 5 to play basketball
ii) if 2 of them are twin ,find the Proab. that the twin play on the same team

My answer:
i) 10C5/2=126
ii) 8C3/126=0.(4)
Check plz More coming

 

[underthesun]

Because, B won the first 2 right?

Technically, there can only be maximum of 7 games left, for example

AAAABBA

now any permutations of the above is the ways which A could win.

Or, B could win instead:

AAAABBB

7 games, permutations of the above is the numbers of way which A or B wins.

numbers of permutations of AAAAABBB is the sum of the amount of the 2 ways above. Because all permutations are equally likely, the chance of winning is the numbers of way that A wins divided by total numbers of way possible.

Uh oh, don't tell me I did something wrong

oh and keep'em coming

 

[ND]

Originally posted by MyLuv
Why 8???
Anyway,here is another 1:

My answer:
i) 10C5/2=126
ii) 8C3/126=0.(4)
Check plz More coming

For ii) i got 4/9:

Whichever team the first twin is on, there's a 4/9 chance the other twin will be on the same team. But this is probability - so i'm probably wrong.

 

[Constip8edSkunk]

Originally posted by underthesun
Because, B won the first 2 right?

Technically, there can only be maximum of 7 games left, for example

...

Uh oh, don't tell me I did something wrong

oh and keep'em coming

Sorry but there are only 5 games played, or so say the question

Originally posted by ND
For ii) i got 4/9:

Whichever team the first twin is on, there's a 4/9 chance the other twin will be on the same team. But this is probability - so i'm probably wrong.

you are right. you both have the same answer. i like this way better though (i did it myluv's way)

 

[underthesun]

Sorry, my post here was wrong. move along, nothing to see

Im writing the supposedly right method now, a friend told me, so im trying..

edit: OK : Here's the scoop.

for (ii), A won the first game.

There can be 5 games played left. 5 games, that A and B plays, and have equal chances.

Hence, it can be AAABB or ABBBB or whatever.

expansion of (A + B)⁵ equals
a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵.

You see, from the expansion above, the ab⁴ parts means B won, because a needs to win twice to win, and b⁵ is the same case. Hence let's move these two out.

Now, the a²b³ combination. Arranging AABBB, there are total 10 arrangements. The arrangements in which the last game is B is the arrangement of A's victory, because A gets 2 wins before B gets 3 wins. There are 4 arrangements of such. Because the arrangements are equally likely, there are 6 of a²b³ coefficient which A wins. The rest of the coefficients, A wins.

So:

a⁵ + 5a⁴b + 10a³b² + 6a²b³

now a = 0.5 and b = 0.5. a⁵ = 1/32.

So, add the numbers all together, and divide by 32.

Happens to be 22/32. Coincidence that it was the same with the previous answer that i posted in this post, and I don't think i obtained it this way...

just realised that skunk already did it

 

[MyLuv]

Two marksmen, A and B, fire simultaneously at a target. If A is twice as likely to hit the target as B, and if the probability that the target does get hit is 1/2 find the probability of A hitting the target. (Answer correct to 3 significant figures

P(A)= 2/3=0.333( wierd )

An orchestra has 2n cellists, n being female and n male. From the 2n cellists a committee of 3 members is formed which contains more females than males.
(i) How many possible committees are there consisting of 2 females and 1 male?
(ii) How many committees are there consisting of 3 females?
(iii) Using the results above,or otherwise,prove that n*nC2 + nC3 = 1/2 2nC3
(iv) If in fact the orchestra has 6 cellists,including Mary and Peter,find the probability that the committee chosen has Mary in it if it is known that Peter has been
chosen.

i) nC1*nC2=n*nC2
ii) nC3
iii) there are 2nC3 way of choosing a committee of 3.Since there are 3 people , No. of female must be either greater or less than No. of male, and No.Female=No.male---> no. of way choosing commitee of 3 and more female is 1/2*2nC3
--> n*nC2+ nC3=1/2 * 2nC3
iv) 2C1/3C2=2/3
Edit:Btw,can some1 check my 1st question plz

 

[MyLuv]

Sum More

There are 12 red, 12 green, 12 yellow and 12 blue cards in a pack. 5 cards are chosen at random. Write down expressions for the probabilitythat:
(i) all 5 cards are red,
(ii) all 5 cards are the same colour,
(iii) at least one card of each colour is chosen.

i)12C5/48C5
ii) 4*12C5/48C5
iii)4!*[12C1]^4/48*47*46*45
-----------------------------------------------
a long one

Consider a pack of 50 playng cards which consist of 5 colours {Yellow,Green,Blue,Indigo and Violet} containing cards numbered from 1 to 10 inclusive,resp ectively. A joker is added to the pack. The joker can stand for any card and when there are equal numbers of diferent cards it takes the value of the higher card. Otherwise,the joker stands for the card which is occurring most often.
e.g. two 4s,t wo 6s and a joker = two 4s and three 6s;
two 4s,one 6,one 8 and a joker = three 4s,one 6 and one 8.
If five cards are dealt to a player,determine the probability (leaving your answer as a fraction in its simplest form) that the player has received:
(i) four 10s.
(ii) any three of one number and any two of another number ,e.g.,three 10s and two 8s.

i) Total way of choosing5=51C5
Having 4 of 10s:
- without Joker=45*5C4
-with Joker=45*5C3
--> Proab=45*(5C4+ 5C3)/51C5=45/156604
ii)
-without Joker=10*5C3*9*5C2
-with Joker=10*5C2*9*5C2*1(coz when there are 2:2 with a Joker will make 3:2 no matter what)
--->Proab=300/39151

 

[underthesun]

Edit:Btw,can some1 check my 1st question plz

Are you talking about your absolute first question?

I think there's something wrong with the question, as the results would be different if there's 3 students in class or 30 students in class, or etc..

 

[MyLuv]

I think it was like the one where proability is constant due to a large number(like the one inExt1 with 2% faslety light )The thing is I cant remember where I saw it