@krum @rubikmath Search Buffon's needle 😃 Open forst link ..wiki
@krum said:seems like integration will come into play, p will depend on the distance of center from line and the angle , do u have explanation?
nope .
@krum said:seems like integration will come into play, p will depend on the distance of center from line and the angle , do u have explanation?
Bhai... Kya baat hai @krum said:seems like integration will come into play, p will depend on the distance of center from line and the angle , do u have explanation?
Explanation..:P
Refer to the 53rd problem..at your own risk..
@soumitrabengeri said:Explanation..www.scribd.com/doc/90521066/27...Refer to the 53rd problem..at your own risk..
Sach me 😛
@rubikmath said:
A table of infinite expanse has inscribed on it a set of parallel lines spaced 2a units apart. A needle of length 2l (less than 2a) is tossed on the table. What is the probability that when it comes to rest , it crosses a line ?OA : 2l/(pi*a)no clue how they got it any explanations ?
instead of posting this in quants thread..u should have posted it in "quantum mechanics" thread....
A coach is training 15 girls. He wants to form 5 lines of 3 forwards each (left-wing, center,and right-wing). Assume that the order of assigning these positions matters. What is the probability that both Ann and May are in the same line?
no OA
no OA
@rubikmath said:A coach is training 15 girls. He wants to form 5 lines of 3 forwards each (left-wing, center,and right-wing). Assume that the order of assigning these positions matters. What is the probability that both Ann and May are in the same line?no OA
see, i'm not taking order of assigning these positions into consideration, because frankly speaking i couldn't comprehend exactly what it means..
i'm going with a simple prob case.
suppose we have three lines:
1 to 5
6 to 10
11 to 15
now out of these, Ann can go in any of the line at any place.
now we are left with 14 positions.
so for May to be in the same line, favorable positions are 4 (the remaining ones in the line in which Ann went.)
hence prob should be 4/14 or 2/7. in this case (i haven't considered order).
hope this can give you an approach for ordered cases.
@rubikmath said:A table of infinite expanse has inscribed on it a set of parallel lines spaced 2a units apart. A needle of length 2l (less than 2a) is tossed on the table. What is the probability that when it comes to rest , it crosses a line ?OA : 2l/(pi*a)no clue how they got it any explanations ?
Make parallel lines with distance (2a) between them.
Now, The needle's center can lie on any point between them, so Probability that it falls at a distance x from one of them is P(falls at x) = 1/2a
Now, that it falls at a distance x, what is the probability that it will cross the line, for that, I introduce the angle A the line makes with the horizontal
so, favorable cases are when A lies between a1 and -a1 and also between (pi - a1 and pi + a1)
where cos(a1) = x/l, hence P(crosses the line)= 2*a1/pi
x varies from 0 to l here.
So, P(the line touches the needle) = P(center of needle falls at x) * P(it touches the line with center at x) = (1/2a)*(2a1/pi) = a1/(pi*a)
Now, we need to integrate it for x lying between 0 and l
int( cos inverse(x/l) / (pi*a) ) over 0 to l => l/(pi*a) [For this used substitution and integration by parts rule]
Now, the probability is same when x is between -l and 0
=> Final probability = 2l/(pi*a)
@rubikmath said:A coach is training 15 girls. He wants to form 5 lines of 3 forwards each (left-wing, center,and right-wing). Assume that the order of assigning these positions matters. What is the probability that both Ann and May are in the same line?no OA
Not so sure, but this is what I did @rubikmath
The total cases are 15! (as the order matters and there are 15 places to be filled, and I assumed that he starts filling from one line to the last line)
Now, for the favorable cases:
1. They are in the same line of 3 (horizontal)
Favorable cases = [(13!)/[((3!)^4)*(4!)] ]*(3!)^5 * 5!
Dividing by total cases gives P = 1/7
2. They are in the same line of 5 (vertical)
Favorable cases = [ (13!)/[(5!)^2 * 3!*2!] ] *(5!)^3 * 3!
Dividing by total cases gives P = 2/7
So, Total = 3/7 ?
@soham2208 said:Not so sure, but this is what I did @rubikmathThe total cases are 15! (as the order matters and there are 15 places to be filled, and I assumed that he starts filling from one line to the last line)Now, for the favorable cases:1. They are in the same line of 3 (horizontal)Favorable cases = [(13!)/[((3!)^4)*(4!)] ]*(3!)^5 * 5!Dividing by total cases gives P = 1/72. They are in the same line of 5 (vertical)Favorable cases = [ (13!)/[(5!)^2 * 3!*2!] ] *(5!)^3 * 3!Dividing by total cases gives P = 2/7So, Total = 3/7 ?
i dont think it is a favourable case : same line of 5(vertical) , as in the question it is , 5 lines of 3.
anyway i don't have the OA .
your answer is right(to me) for the cases you have considered .
anyway i don't have the OA .
your answer is right(to me) for the cases you have considered .
@rubikmath said:i dont think it is a favourable case : same line of 5(vertical) , as in the question it is , 5linesof 3.
anyway i don't have the OA .
your answer is right(to me) for the cases you have considered .
5 vertical lines are not considered I would like to try an alternate solution which comes to 3/91
toltal ways = 15!
favourable = 5c1 *13c1*3!*12c2*12!
probaility = 1/7 ???
edited calculation mistake
@techsurge said:getting 2/91arange these in 15! waysFavourable arrangement = 5c1 *13c1 *3! *(12c12 *12!) probability = 6/13*14=3/91
dude you are right. But,check the calculations 
. Answer by your logic would be 1/7 only 
. Answer by your logic would be 1/7 only @rubikmath said:dude you are right. But,check the calculations . Answer by your logic would be 1/7 only
sahi kaha 1/7
:banghead:
thats why i lose out cause of my silly calculation mistakes
@rubikmath said:A coach is training 15 girls. He wants to form 5 lines of 3 forwards each (left-wing, center,and right-wing). Assume that the order of assigning these positions matters. What is the probability that both Ann and May are in the same line?no OA
Select any five row, fill that row with ann and may and arrange it. P(A)=1/15, P(M)=1/14
5c1*(1/15)*(1/14)*3! = 1/7
5c1*(1/15)*(1/14)*3! = 1/7
X and Y are natural nos. and X is odd Find the no. of solutions of : 1/X = 1/18 - 3/Y
@Torque024 said:X and Y are natural nos. and X is odd
1/X + 3/Y = 1/18
18Y + 54X = XY
(18-X)*Y + 54X - 54*18 = -54*18
(18-X)(Y - 54) = -54*18
=> (X-18)*(Y-54) = 54*18
X is odd -> X - 18 = odd
54*18 = 2^2 * 3^5
X - 18 = 1/3/3^2/3^3/3^4 => 5 solutions ?
@Torque024 said:X and Y are natural nos. and X is odd
(x-18)(y-54)=18*54
=2^2*3^5
(x-18)=1,3,9,27,81
5 values
=2^2*3^5
(x-18)=1,3,9,27,81
5 values