@naveenkrs but isn't a lap defined (in this case as it is not a circular track)---as going from 0m to 50m as one lap and coming back from 50m to 0m point as the second and so on..?
@pratskool said:total number of outcomes = 144no. of outcomes with no head = 11 head = 102 head = 36 (selecting 2 gaps among 9 gaps)3 head = 56 (selecting 3 among 8 gaps)4 head = 15 (selecting 4 among 6 gap)5 head = 2so 128/1024 ??
Bhai
6T 4H = 7c4 = 35..
5T 5H = 6c5 = 6
so u total cases = 144....
got it now
max. 5 heads can occur and min. 2 heads as per given condition
hence
5 heads occur = 6c5
4 heads occur = 7c4
3 heads occur =8c3
2 heads occur = 9c2
total ways = 128 therefore 128/1024= 1/8
@shattereddream said:why answer is not correct whats the OA then ?
@Logrhythm said:how?
go through @pratscool's solution.
@The_Loser said:go through @pratscool's solution.
my solution is wrong... did it in haste... @Logrhythm is correct.... it should be 144/1024... if only it was mentioned a minimum of 2 heads, then it would have been 1/8
@The_Loser said:in a room ther are 7 persons. the chances that two of them were born on the same day of week.?
prob= 7*7C2* (1*1*6*5*4*3*2)/(7^7) = 2160/(7^5) (Birthday problem)
@The_Loser said:in a room ther are 7 persons. the chances that two of them were born on the same day of week.?
choosing 2 ppl -> 7c2
choosing that day of the week -> 7c1
1st person has -> 1/7 option
2nd person has -> 1/7 option
3rd person has -> 6/7
4th -> 5/7
5th -> 4/7
6th -> 3/7
7th -> 2/7
so prob = 21*7*6*5*4*3*2/7^7 = 2160/7^5
@The_Loser said:in a room ther are 7 persons. the chances that two of them were born on the same day of week.?
since exactly is not mentioned... 1stly total no. of cases = 7^7
secondly, none of the two being born on same day number of cases is 7!
so required probability = 1 - 7!/(7^7)
100 identical coins of each with probability P of showing up heads are tossed once. If 01/21
49/101
50/101
51/101
How many right triangles with integer sides exist such that 1215 is one of the smaller sides?
1. 8
2. 12
3. 10
4. 16
1. 8
2. 12
3. 10
4. 16
@The_Loser said:100 identical coins of each with probability P of showing up heads are tossed once. If 01/2149/10150/10151/101
50/101??
solve the equation 100C50* (1-P)^50*p^50 = 100C51 * (1-p)^49*p^51
Q:
N, THE SET OF NETURAL NO IS PARTITIONED INTO SUBSET S1=(1) , S2 =(2,3) ,S3=(4,5,6) ,S4= (7,8,9,10) AND SO ON . THE SUM OF THE ELEMENTS OF THE SUBSET S50 IS: ?????
FIND THE SUM OF THE FIRST 16 TERMS OF THE SEQUENCE
2,2,5,4,8,8,11,16....?????
2,2,5,4,8,8,11,16....?????
@pavimai said:How many right triangles with integer sides exist such that 1215 is one of the smaller sides?1. 82. 123. 104. 16
whats OA
@uditultimate said:Q:N, THE SET OF NETURAL NO IS PARTITIONED INTO SUBSET S1=(1) , S2 =(2,3) ,S3=(4,5,6) ,S4= (7,8,9,10) AND SO ON . THE SUM OF THE ELEMENTS OF THE SUBSET S50 IS: ?????
62525 ?
@uditultimate said:Q:N, THE SET OF NETURAL NO IS PARTITIONED INTO SUBSET S1=(1) , S2 =(2,3) ,S3=(4,5,6) ,S4= (7,8,9,10) AND SO ON . THE SUM OF THE ELEMENTS OF THE SUBSET S50 IS: ?????
62550
@pavimai said:How many right triangles with integer sides exist such that 1215 is one of the smaller sides?1. 82. 123. 104. 16
1215 = 3^4*5
pythagorean triplet -> 3,4,5..
there are 4 3's but the smaller sides can be used interchangeably. hence, 4*2 = 8 such triangles??
Not at all sure abt this.. 
@uditultimate said:FIND THE SUM OF THE FIRST 16 TERMS OF THE SEQUENCE 2,2,5,4,8,8,11,16....?????
610
@pavimai said:How many right triangles with integer sides exist such that 1215 is one of the smaller sides?1. 82. 123. 104. 16
answer si 12