- View Source
let us divide the circle in to 2 semicircles.the prob for a selected point to be on one of the semicircle is equal to the prob of being on the other.so prob for 8 selected points to be on any one of semicircles = 2*1/256 = 1/128.

On Thu, 24 Jun 2004 josephite_shashi wrote :>hii friends,

>i have a problem? plzz help me to solve it.

>8 points are selected at random on a circle, what is the probability

>they lie on a semicircle?

>regards

>shashi

>Ascent Education

>An IIM Alumni Venture

>Class for CAT, XAT, GRE, GMAT

>http://www.ascenteducation.com

>Archives of past CAT questions can be viewed at

>http://www.ascenteducation.com/india-mba/iim/cat/questionbank/questionbank.shtml

>Yahoo! Groups Sponsor

>ADVERTISEMENT

>Yahoo! Groups Links

>To visit your group on the web, go to:

>http://groups.yahoo.com/group/ascent4cat/

>

>

>To unsubscribe from this group, send an email to:

>ascent4cat-unsubscribe@yahoogroups.com

>

>

>Your use of Yahoo! Groups is subject to the

>Yahoo! Terms of Service

>.

>

- View Sourcehi rahul

it should be minimum outside surface "area"

the surface area for a cylindrical bucket with one side open is:

pi*(r^2)[bottom]+2*pi*r*h[curved surface] say S

the volume

pi*r^2*h=30*1000 c.c [given]

this gives h=30000/(pi*r^2) cm

substituting in eq for area gives:

S= pi*r^2+ 60000/r cm sq

for this to be min dS/dr=0 [calculas funda] and d2S/dr2 be +ve

which in this case is 6000*pi so

which gives r^3=30000/pi;

which gives

S= 9000*(pi/30)^(1/3)

S=4242.14 cm sq

--- In ascent4cat@yahoogroups.com, <rahul.soni@w...> wrote:

> hi,

> solve this question....

>

>

> Calculate the minimum outside surface of a cylindrical bucket

with an open upper side and which can hold 30 liters of water.

>

> The Question: What is the minimum outside surface?

>

>

> Thanks & Regards,

> Rahul Soni - View SourceI feel that the probability should be (1/2)^8 i.e. 1/256. Since the two semi-circles are imaginary and so cannot be differntiated as such and so we shuld not multiply by 2.
wrote:*hai crux <dgb_3@...>*let us divide the circle in to 2 semicircles.the prob for a selected point to be on one of the semicircle is equal to the prob of being on the other.so prob for 8 selected points to be on any one of semicircles = 2*1/256 = 1/128.

On Thu, 24 Jun 2004 josephite_shashi wrote :

>hii friends,

>i have a problem? plzz help me to solve it.

>8 points are selected at random on a circle, what is the probability

>they lie on a semicircle?

>regards

>shashi

>Ascent Education

>An IIM Alumni Venture

>Class for CAT, XAT, GRE, GMAT

>http://www.ascenteducation.com

>Archives of past CAT questions can be viewed at

>http://www.ascenteducation.com/india-mba/iim/cat/questionbank/questionbank.shtml

>Yahoo! Groups Sponsor

>ADVERTISEMENT

>Yahoo! Groups Links

>To visit your group on the web, go to:

>http://groups.yahoo.com/group/ascent4cat/

>

>

>To unsubscribe from this group, send an email to:

>ascent4cat-unsubscribe@yahoogroups.com

>

>

>Your use of Yahoo! Groups is subject to the

>Yahoo! Terms of Service

>.

>

`Ascent Education`

An IIM Alumni Venture

Class for CAT, XAT, GRE, GMAT

http://www.ascenteducation.com

Archives of past CAT questions can be viewed at http://www.ascenteducation.com/india-mba/iim/cat/questionbank/questionbank.shtml

Do you Yahoo!?

Yahoo! Mail Address AutoComplete - You start. We finish.