What is the probability of selecting 2 points on a straight line of length 60 cm such that the distance between the 2 does not exceed 20 cm?
Options
1) 1/4
2) 2/3
3) 4/9
4) 5/9
OA 5/9
@nick_baba said:which post buddy..though i have edited one ...neways..try to solve it..n tell me the logic
no logic as such..just tried for some values
@krum said:What is the probability of selecting 2 points on a straight line of length 60 cm such that the distance between the 2 does not exceed 20 cm?Options 1) 1/42) 2/33) 4/94) 5/9OA 5/9
Probability that one point is within 20 m of other = 20/60 = 1/3
Probability that one is not in the required range = 2/3
Similarly for the other point, it will be 2/3
So, probability that they are not within 20 m of each other is (2/3)*(2/3) = 4/9
Hence probability of choosing 2 points not more than 20 m apart from each other is 1 - 4/9 = 5/9
Probability that one is not in the required range = 2/3
Similarly for the other point, it will be 2/3
So, probability that they are not within 20 m of each other is (2/3)*(2/3) = 4/9
Hence probability of choosing 2 points not more than 20 m apart from each other is 1 - 4/9 = 5/9
Please correct me if i am wrong.
@soumitrabengeri said:Probability that one point is within 20 m of other = 20/60 = 1/3Probability that one is not in the required range = 2/3Similarly for the other point, it will be 2/3So, probability that they are not within 20 m of each other is (2/3)*(2/3) = 4/9Hence probability of choosing 2 points not more than 20 m apart from each other is 1 - 4/9 = 5/9Please correct me if i am wrong.
A------20m--------B------20m--------C------20m------D
say i chose B, p(other point within 20m)=2/3
here probability is not fixed but changes linearly
say i chose B, p(other point within 20m)=2/3
here probability is not fixed but changes linearly
@soumitrabengeri said:Probability that one point is within 20 m of other = 20/60 = 1/3Probability that one is not in the required range = 2/3Similarly for the other point, it will be 2/3So, probability that they are not within 20 m of each other is (2/3)*(2/3) = 4/9Hence probability of choosing 2 points not more than 20 m apart from each other is 1 - 4/9 = 5/9Please correct me if i am wrong.
your explanation luks right but please help me with this justification..
why we have to consider not possible cases n subtract them from 1..
why cudn't it be like prob of one being in the 20cm vicinity of the other=1/3 similarly for the other..hence why not 1/9??
@krum said:A------20m--------B------20m--------C------20m------Dsay i chose B, p(other point within 20m)=2/3here probability is not fixed but changes linearly
bhai is thi ques of similar class to the one that was discussed in 2012 thread k wat z prob for meeting of 2 frnz within one hour if one waits for the other for not more than 15 mins???
@krum said:What is the probability of selecting 2 points on a straight line of length 60 cm such that the distance between the 2 does not exceed 20 cm?Options 1) 1/42) 2/33) 4/94) 5/9OA 5/9
let us first consider a line on x- axis starting from coordinates (0,0) to (60,0)
let x and y be the x-coordinates of the two points we have to choose.
now according to the given condition we want |x-y| less than equal to 20
now to choose two points on the given line
no. of possible ways =61*61 ( as x,y can vary from [0,60]
now for no. of favorable outcomes,lets start making cases
for |x-y| =0, 61 cases x=y
for |x-y| =1 ,60 cases for x>y and 60 cases for x
.
.
.
.
.
for |x-y| =20 ,2*41 (41 for each)
so total favorable cases=61 + 2*(41 +42 +........60)=2081
so probability=2081/61*61=5/9
@soumitrabengeri said:Probability that one point is within 20 m of other = 20/60 = 1/3Probability that one is not in the required range = 2/3Similarly for the other point, it will be 2/3So, probability that they are not within 20 m of each other is (2/3)*(2/3) = 4/9Hence probability of choosing 2 points not more than 20 m apart from each other is 1 - 4/9 = 5/9Please correct me if i am wrong.
@nick_baba said:bhai is thi ques of similar class to the one that was discussed in 2012 thread k wat z prob for meeting of 2 frnz within one hour if one waits for the other for not more than 15 mins???
@busar005 said:let us first consider a line on x- axis starting from coordinates (0,0) to (60,0)let x and y be the x-coordinates of the two points we have to choose.now according to the given condition we want |x-y| less than equal to 20now to choose two points on the given lineno. of possible ways =61*61 ( as x,y can vary from [0,60]now for no. of favorable outcomes,lets start making casesfor |x-y| =0, 61 cases x=yfor |x-y| =1 ,60 cases for x>y and 60 cases for x......for |x-y| =20 ,2*41 (41 for each)so total favorable cases=61 + 2*(41 +42 +........60)=2081so probability=2081/61*61=5/9
for the first 20cm the probability increases linearly from 1/3 to 2/3
it remains constant for next 20cm and decreases linearly from 2/3 to 1/3 for last 20cm
so p=1/3*(1/3+2/3)/2+1/3*2/3+1/3*(1/3+2/3)/2
=>1/6+2/9+1/6
=>(3+4+3)/18
=>10/18
=>5/9
it remains constant for next 20cm and decreases linearly from 2/3 to 1/3 for last 20cm
so p=1/3*(1/3+2/3)/2+1/3*2/3+1/3*(1/3+2/3)/2
=>1/6+2/9+1/6
=>(3+4+3)/18
=>10/18
=>5/9
Let p1, p2, p3 and p4 be distinct prime numbers satisfying :
2p1 + 3p2 + 5p3 + 7p4 = 162
11p1 + 7p2 + 5p3 + 4p4 = 162
If p1p2p3p4 = k find the number of possible values of k?
a) 0
2p1 + 3p2 + 5p3 + 7p4 = 162
11p1 + 7p2 + 5p3 + 4p4 = 162
If p1p2p3p4 = k find the number of possible values of k?
a) 0
b) 1
c) 2
d) 3
@krum said:for the first 20cm the probability increases linearly from 1/3 to 2/3it remains constant for next 20cm and decreases linearly from 2/3 to 1/3 for last 20cmso p=1/3*(1/3+2/3)/2+1/3*2/3+1/3*(1/3+2/3)/2=>1/6+2/9+1/6=>(3+4+3)/18=>10/18=>5/9
thanks for the explanation..did not realize the fact that the prob changes linearly and is not fixed.
@nick_baba said:an easy one...For how many primes p is p^2 + 3p ˆ' 1 also prime?a) 0 b) 1 c) 2 d) 3 koi mast sa general sol batao..option dekh ke mat karna..
(p+1)(p-1)+3p this number is always divisible by 3 so ans is zero
@krum said:for the first 20cm the probability increases linearly from 1/3 to 2/3it remains constant for next 20cm and decreases linearly from 2/3 to 1/3 for last 20cmso p=1/3*(1/3+2/3)/2+1/3*2/3+1/3*(1/3+2/3)/2=>1/6+2/9+1/6=>(3+4+3)/18=>10/18=>5/9
can you plzz explain me sir how the the probability is increasing in first 20 cm and it is 2/3 in next...??
by the way,great method..
@krum said:for the first 20cm the probability increases linearly from 1/3 to 2/3it remains constant for next 20cm and decreases linearly from 2/3 to 1/3 for last 20cmso p=1/3*(1/3+2/3)/2+1/3*2/3+1/3*(1/3+2/3)/2=>1/6+2/9+1/6=>(3+4+3)/18=>10/18=>5/9
whooosshhhh.......sar ke upar se gaya .. guruji thoda vistar se(ho sake to diagram )..jaise ATDH sir smjhate hai..
@HeCtEr said:smjha nhi y b?
http://pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-c-t-88456/p-3611604/r-4027641?page=42
find the post#825 by krum..its self explanatory...plus if u put p=3..u'll get a prime number .i.e. 17
There are two ants on opposite corners of a cube. On each move,
they can travel along an edge to an adjacent vertex. If the probability that
they both return to their starting position after 4 moves is m/n , where m and
n are relatively prime integers, find m+n. (NOTE:They do not stop if they
collide.)
a) 17
they can travel along an edge to an adjacent vertex. If the probability that
they both return to their starting position after 4 moves is m/n , where m and
n are relatively prime integers, find m+n. (NOTE:They do not stop if they
collide.)
a) 17
b) 65
c) 73
d) 85
@nick_baba said:pagalguy.com/forums/quantitati...find the post#825 by krum..its self explanatory...plus if u put p=3..u'll get a prime number .i.e. 17
got it . p=3 will be the only value though.
@nick_baba said:Let p1, p2, p3 and p4 be distinct prime numbers satisfying :2p1 + 3p2 + 5p3 + 7p4 = 16211p1 + 7p2 + 5p3 + 4p4 = 162If p1p2p3p4 = k find the number of possible values of k?a) 0 b) 1 c) 2 d) 3
0?