[jigar_p_civil]

Originally Posted by itsromiljain

Question:-Two friends decide to meet at a place between 4-5 o clock..
anyone who comes there first waits for the other for 10 mins and then leaves..
whats the probability that they will meet??

Please solve it and tell me the approach.

still this question didnt solved in any thread or any site(totalgadha.com & cat4mba.com)

may be better if u upolad option plzzzzzzzzzzzzzzzzz...........

[itsromiljain]

Originally Posted by jigar_p_civil

still this question didnt solved in any thread or any site(totalgadha.com & cat4mba.com)

may be better if u upolad option plzzzzzzzzzzzzzzzzz...........

Ya I know abt it. @TG very few ppl posted the answer dats vy i posted it here.
I dont have the options if some body have solved it nd know the answer then tell me.

[ssaaggaar]

Question1.A candidate is selected for three interview for three posts.For first post there are5 candidates,for second post there are 8 candidates and for third position there are 7 candidates.What are the chances for his getting atleast one post?

[jigar_p_civil]

Originally Posted by ssaaggaar

Question1.A candidate is selected for three interview for three posts.For first post there are5 candidates,for second post there are 8 candidates and for third position there are 7 candidates.What are the chances for his getting atleast one post?

ans is 2/5?

[shashank3012]

Originally Posted by ssaaggaar

Question1.A candidate is selected for three interview for three posts.For first post there are5 candidates,for second post there are 8 candidates and for third position there are 7 candidates.What are the chances for his getting atleast one post?

selected for 1 post only=>42/210+28/210+24/210=94/210

selected for 2 posts=>17/210

selected for 3 posts=>1/210

probability=>112/210

so,8/15.

what is the answer??

[jigar_p_civil]

Originally Posted by shashank3012

selected for 1 post only=>42/210+28/210+24/210=94/210

selected for 2 posts=>17/210

selected for 3 posts=>1/210

probability=>112/210

so,8/15.

what is the answer??

not select for any post (4/5)*(7/8 )*(6/7)=3/5

select for atleast one post=2/5

[shashank3012]

Originally Posted by jigar_p_civil

not select for any post (4/5)*(7/8 )*(6/7)=3/5

select for atleast one post=2/5

ok,by marginalization

i considered probability,that he would be selected as 1/5 and not selected as 7/8 and 6/7 and so on......considered all the cases.

[Derrida]

Originally Posted by itsromiljain

Question:-Two friends decide to meet at a place between 4-5 o clock..
anyone who comes there first waits for the other for 10 mins and then leaves..
whats the probability that they will meet??

Please solve it and tell me the approach.

Let's say one guy comes x minutes after 4 and the other guy comes y minutes after 4. Clearly 0<=x,y<=60. Consider a square bound by the lines x=0, y=0, x= 60 and y = 60. This square is obviously collection of all the points (x,y) which signify the time of arrival of the two friends.

The two friends can come at any time in any order though they will meet only if |x-y|<=10. Let's draw the lines |x-y|<=10, Thus the required probability is the ratio of area of square contained between the two lines to the total area of the square = 11/36

P.S.: tried attaching the diagram a hazaar baar, still upload failed :(.

[nsvikin]

Here's one i dont understand:

A man has 10% probability of hitting a target. How many shots should he fire so that his probability increases above 50%?

Options are 9,11,7,5

Answer is 7... But how?