@sujamait said:ohh bhari bhari Q daal rhe ho a .............. ba+b=13/2b=a+(n-1)dd=b-a/n-1let there be p (even) means,p/2(2(a+d) + (p-1)*d) - p = 1p nikal hee jayega isse..pen nhn hai yar..
Pen kyun bhool gaye sir 
@ScareCrow28 said:Pen kyun bhool gaye sir
koi utha ke le gaya yar
In a particular turn of a game, a person can win 2 rupees with a probability of 0.75 and otherwise, lose 1 rupee with a probability of 0.25. What is the probability that after 7 turns a person would have won exactly two rupees? (Assume he had more than 7 rupees to start with)
OA sarv sahmati se!!
In a particular turn of a game, a person can win 2 rupees with a probability of 0.75 and otherwise, lose 1 rupee with a probability of 0.25. What is the probability that after 7 turns a person would have won exactly two rupees? (Assume he had more than 7 rupees to start with)
OA sarv sahmati se!!
@sujamait said:Yeah sahi hain sab log
Yeah after doing completely got 998 😛 😛
moral : do not trust your memory 😛 😛
@sujamait said:Yeah sahi hain sab log
The answer is 997 it seems. Refer this:-> http://testfunda.com/examprep/learningresources/qod/question-of-the-day.htm?assetid=a467100b-c6dc-4c5c-99e7-05152ae1ca97
@sujamait said:koi utha ke le gaya yar In a particular turn of a game, a person can win 2 rupees with a probability of 0.75 and otherwise, lose 1 rupee with a probability of 0.25. What is the probability that after 7 turns a person would have won exactly two rupees? (Assume he had more than 7 rupees to start with)OA sarv sahmati se!!
2a - b = 2
a + b = 7
a = 3, b = 4
a = 3, b = 4
probability = C(7, 3)*(3/4)^3 * (1/4)^4 = 945/16384
@sujamait said:koi utha ke le gaya yar In a particular turn of a game, a person can win 2 rupees with a probability of 0.75 and otherwise, lose 1 rupee with a probability of 0.25. What is the probability that after 7 turns a person would have won exactly two rupees? (Assume he had more than 7 rupees to start with)OA sarv sahmati se!!
945/16384??????
@sujamait said:koi utha ke le gaya yar In a particular turn of a game, a person can win 2 rupees with a probability of 0.75 and otherwise, lose 1 rupee with a probability of 0.25. What is the probability that after 7 turns a person would have won exactly two rupees? (Assume he had more than 7 rupees to start with)OA sarv sahmati se!!
2x - y = 2
Also, x+y=7
=> x= 3 and y= 4
Probability: 7C3*(3/4)^3*(1/4)^4 ...
Kya ye sahi hai sir?
Kya ye sahi hai sir? @ScareCrow28 said:Between two numbers whose sum is 6 ˝ an even number of arithmetic means is inserted;the sum of these means exceeds their number by unity. How many means are there?(1) 12 (2) 6 (3) 24 (4) None
Anyone for this?
@SCARECROW28
Answer : none of these
n+1+6.5 = (n+ 2/2)*(6.5)
@ScareCrow28 said:Anyone for this?
a + b = 13/2
a A1 A2......A2n b
n* 13/2 = 2n + 1
13n = 4n + 2
9n = 2
n = 2/9
2n = 4/9 ? :O
@rkshtsurana said:a + b = 13/2a A1 A2......A2n bn* 13/2 = 2n + 113n = 4n + 29n = 2 n = 2/9 2n = 4/9 ?
me too getting d same.....
@rkshtsurana said:a + b = 13/2a A1 A2......A2n bn* 13/2 = 2n + 113n = 4n + 29n = 2 n = 2/9 2n = 4/9 ?
y did u do n*13/2 = 2n+1?@sujamait said:
The function f(n) is defined on the set of integers. f(n) satisfies the given conditions;f(n) = n €“ 3 if n ‰Ľ 1000= f(f(n + 5)) if n Find f(84).OPTIONS1) 100 2) 84 3) 998 4) 997
f(84) = f(f(89)) = f(f(f(94))) = .... = f(f(....f(1004))...), where f is applied 185 times
f(1004) = 1001
f(f(1004)) = 998
f(f(f(1004))) = 997
f(1004) = 1001
f(f(1004)) = 998
f(f(f(1004))) = 997
f(f(f(f(1004)))) = 998, so it follows a cycle of 2 (barring first one)
185 is odd, so 997 will be the answer
@krum
@krum said:35?
((7-1)^83+(7+1)*83)mod49
=(83*7+83*7)mod49
=1162mod49
=35
can you explain (83*7 + 83*7) mod 49 step, how u reach at this step?
@YouMadFellow said:f(n) = (7-1)^n + (7+1)^nn = 83 -> f(83) = 2* 83C1(7) + higher terms (terms containing 49)f(83) mod 49 = 35
can you explain in detail this step :-
f(83) = 2* 83C1(7) + higher terms (terms containing 49)
f(83) mod 49 = 35
@ScareCrow28 said:y did u do n*13/2 = 2n+1?
Sum of AM = 13/2 * 2n/2
sum of a term from last and start is same for an AP..so a+ b = 13/2
then A1+ A2n = 13/2...so total 2n AM..taking two at a time so multiplied by n
@eni said:@krum can you explain (83*7 + 83*7) mod 49 step, how u reach at this step?
(7 - 1)^83 + (7 + 1)^83
= 2{7^83 + C(83, 2)*7^81 + C(83, 4)*7^79 + ..... + C(83, 82)*7}
except the last term, all have exponent of 7 more than 2 (means divisible by 49)
= 2{7^83 + C(83, 2)*7^81 + C(83, 4)*7^79 + ..... + C(83, 82)*7}
except the last term, all have exponent of 7 more than 2 (means divisible by 49)
so, we are left with 2*83*7, (so just find the remainder by this)
@rkshtsurana said:Sum of AM = 13/2 * 2n/2sum of a term from last and start is same for an AP..so a+ b = 13/2then A1+ A2n = 13/2...so total 2n AM..taking two at a time so multiplied by n
ans?
@sujamait said:koi utha ke le gaya yar In a particular turn of a game, a person can win 2 rupees with a probability of 0.75 and otherwise, lose 1 rupee with a probability of 0.25. What is the probability that after 7 turns a person would have won exactly two rupees? (Assume he had more than 7 rupees to start with)OA sarv sahmati se!!
3^3 / 4^6 x 7!/3!4!
@eni said:can you explain in detail this step :- f(83) = 2* 83C1(7) + higher terms (terms containing 49)f(83) mod 49 = 35
f(83) = (7-1)^83 + (7+1)^83 = 2*(83C1*7 + 83C3*7^3+...+ 83C83*7^83) .
Note that, barring the 1st term every term is divisible by 49,
so f(83)mod49= 2*83*7mod49 (as rest of the terms is divisible by 49) = 35.