@grkkrg said:16:7?A's chance = 1 - 6C3/9C3 = 16/21 B's chance = 1 - 2/3 = 1/3Ratio = (16/21)/(1/3) = 16/7
Not Fair!! 
@albiesriram said:3:1??
@grkkrg said:16:7?A's chance = 1 - 6C3/9C3 = 16/21 B's chance = 1 - 2/3 = 1/3Ratio = (16/21)/(1/3) = 16/7
@ScareCrow28 said:No of ways of A winning :1 prize : 3C1 * 6C2 2 prizes: 3C2 * 6C13 prizes: 3C3 * 6C0Total no of ways = 9C3Prob = 3C1*6C2 + 3C2*6C1 + 3C3*6C0 / 9C3 = 16/21No of ways of B winning:1 Prize : 1C1 * 2C0Total ways = 3C1Prob = 1C1*2C0/3C1 = 1/3Ratio = 16*3/21 = 16:7 ??
OA-16:7
@pankaj1988 said:'A' has three share in a lottery in which there are 3 prizes and 6 blanks' 'B' has one share in a lottery in which there is 1 prize and 2 blanks. What is ratio of A's chance of winning a prizes to B's chance of winning a prizes.
16:7?
A, B and C in order, toss a coin. The one who gets a head first wins. Find their respective probabilities of winning.
@pankaj1988 said:A, B and C in order, toss a coin. The one who gets a head first wins. Find their respective probabilities of winning.
(a,b,c) = (4/7 , 2/7 , 1/7 )
@pankaj1988 said:A, B and C in order, toss a coin. The one who gets a head first wins. Find their respective probabilities of winning.
A = 4/7
B = 2/7
C = 1/7
A = 1/2 + 1/2*1/2*1/2*1/2 + (1/2)^7 + ... = (1/2)/(7/8) = 4/7
B = 1/2*1/2 + (1/2)^5 + (1/2)^8 + ... = (1/4)/(7/8) = 2/7
C = (1/2)^3 + (1/2)^6 + (1/2)^9 + ... = (1/8)/(7/8) = 1/7
B = 2/7
C = 1/7
A = 1/2 + 1/2*1/2*1/2*1/2 + (1/2)^7 + ... = (1/2)/(7/8) = 4/7
B = 1/2*1/2 + (1/2)^5 + (1/2)^8 + ... = (1/4)/(7/8) = 2/7
C = (1/2)^3 + (1/2)^6 + (1/2)^9 + ... = (1/8)/(7/8) = 1/7
@pankaj1988 said:A, B and C in order, toss a coin. The one who gets a head first wins. Find their respective probabilities of winning.
4/7;2/7;1/7?
@pankaj1988 said:A, B and C in order, toss a coin. The one who gets a head first wins. Find their respective probabilities of winning.
A : 1/2 + 1/2^4 + 1/2^7+ .... = 4/7
B: 1/2^2 + 1/2^5 + ... = 2/7
C: 1/2^3+1/2^6+... = 1/7
Typing speed sucks!!
@Brooklyn said:(a,b,c) = (4/7 , 2/7 , 1/7 )
@ScareCrow28 said:A : 1/2 + 1/2^4 + 1/2^7+ .... = 4/7 B: 1/2^2 + 1/2^5 + ... = 2/7 C: 1/2^3+1/2^6+... = 1/7
@rubikmath said:4/7;2/7;1/7?
@grkkrg said:A = 4/7B = 2/7C = 1/7A = 1/2 + 1/2*1/2*1/2*1/2 + (1/2)^7 + ... = (1/2)/(7/8) = 4/7B = 1/2*1/2 + (1/2)^5 + (1/2)^8 + ... = (1/4)/(7/8) = 2/7C = (1/2)^3 + (1/2)^6 + (1/2)^9 + ... = (1/8)/(7/8) = 1/7
OA : SAME AS urs
@mailtoankit said:kaise yaar?
When finding remainder..If you remove a factor..then you have to add(multiply) it at the end.
You removed 10^6..Now is the time to add(multiply) it!! :D
Ravan has hundred sons.He takes every 10 sons to a jungle for hunting with no ten being repeated.Let X be the number of times that Ravan goes to the jungle and Y be the number of times that each son goes to the jungle.
Find X-Y.
Find X-Y.
A bag contains 3 Red, 2 Green and 4 Yellow balls. A ball is drawn and if it is Red or Yellow it is replaced otherwise not. Find the probability that second ball drawn is a yellow ball.
@pankaj1988 said:A bag contains 3 Red, 2 Green and 4 Yellow balls. A ball is drawn and if it is Red or Yellow it is replaced otherwise not. Find the probability that second ball drawn is a yellow ball.
37/81?
@19rsb said:Ravan has hundred sons.He takes every 10 sons to a jungle for hunting with no ten being repeated.Let X be the number of times that Ravan goes to the jungle and Y be the number of times that each son goes to the jungle.Find X-Y.
9? :P
X = 10
Y = 1
X-Y = 9 :P
X = 10
Y = 1
X-Y = 9 :P
@pankaj1988 said:A bag contains 3 Red, 2 Green and 4 Yellow balls. A ball is drawn and if it is Red or Yellow it is replaced otherwise not. Find the probability that second ball drawn is a yellow ball.
37/81