@grkkrg hmm kk but i thought he asked hw many people before him not days but nyhow whats the answree?
You are playing Bridge and with a well shuffled pack of 52 cards,what is the probability that you are dealt a perfect hand(13 of one suit)? (In Bridge , there are 4 players and each of them is dealt 13 cards )
@kingsleyx said:Let a , b, c, d be positive integers. Then the product (a-b) (a-c) (a-d) (b-c) (b-d) (c-d) is definitely divisible by a)2 b)3 c)8 d)12.select all options which satisfy it !!
a and b ??
EDIT: 12 also..... so a, b and d
@rubikmath said:which is more likely ? i)atleast 1 six when 6 dice are rolled ii)atleast 2 sixes when 12 dice are rolled iii)atleast 3 sixes when 18 dice are rolled ?
(1-(5/6)^6) > (1-[(5/6)^12+(12c1*1/6*(5/6)^11) > (1-[(5/6)^18+(18c1*1/6*(5/6)^17)+(18c2*1/6*1/6*(5/6)^16)])
1) atleast 1 six when 6 dice are rolled is more likely
1) atleast 1 six when 6 dice are rolled is more likely
@ananthchief said:The balance of a trader weighs 10% less than what it should be. Still the trader marks up his goods to get an overall profit of 20%. What is the markup on the cost price.
If we take CP as 100 then MP is 108... so its 8%...
@rubikmath said:You are playing Bridge and with a well shuffled pack of 52 cards,what is the probability that you are dealt a perfect hand(13 of one suit)? (In Bridge , there are 4 players and each of them is dealt 13 cards )
Not much sure though...
52*12!/52C13 ???
@rubikmath said:You are playing Bridge and with a well shuffled pack of 52 cards,what is the probability that you are dealt a perfect hand(13 of one suit)? (In Bridge , there are 4 players and each of them is dealt 13 cards )
4/52c13?
@rubikmath said:You are playing Bridge and with a well shuffled pack of 52 cards,what is the probability that you are dealt a perfect hand(13 of one suit)? (In Bridge , there are 4 players and each of them is dealt 13 cards )
4/52C13
@rubikmath said:how did you get 52*12! ??
4/52c13= 4*13!*39!/52! = 4*13*12!*39!/52! = 52*12!*39!/52!
What is OA BTW?
What is OA BTW?
@rubikmath said:how did you get 52*12! ??
i did it like this...
for first card u can choose and card out of 52... so the no of options are 52
Now for the second card you are restricted to the same suit, that of the first card.
Now u hv 12 cards of that suit which can be choosen in any order so 12!
thats how 52*12!...
i dont think so its correct. wats the OA?
@ScareCrow28 said:Its more than 1 na?
I guess the prob is in the denominator.... not sure though.. it shud be 52P13.
lets see wats the OA is.
@rubikmath said:You are playing Bridge and with a well shuffled pack of 52 cards,what is the probability that you are dealt a perfect hand(13 of one suit)? (In Bridge , there are 4 players and each of them is dealt 13 cards )
tot case=selecting 13 frm 52
fav case=selecting 1 suit frm 4
prob=4c1/52c13?
fav case=selecting 1 suit frm 4
prob=4c1/52c13?
@rubikmath said:OA - 4/52C13
its the same ans i had given...
just open it...
i made mistake in base...
it shud rather be... 52*12!/ 52P13
@maddy2807 said:its the same ans i had given...just open it... i made mistake in base...it shud rather be... 52*12!/ 52P13
Ib to sahi hai :D
@maddy2807 said:i did it like this...for first card u can choose and card out of 52... so the no of options are 52Now for the second card you are restricted to the same suit, that of the first card.Now u hv 12 cards of that suit which can be choosen in any order so 12!thats how 52*12!...i dont think so its correct. wats the OA?
you are mentioning ' in any order ' .but 12! is ordering .
so your answer, divide it by 13! to take care of that ordering .
so your answer, divide it by 13! to take care of that ordering .
@rubikmath said:You are playing Bridge and with a well shuffled pack of 52 cards,what is the probability that you are dealt a perfect hand(13 of one suit)? (In Bridge , there are 4 players and each of them is dealt 13 cards )
4/52c13
first i we select 13 cards of same suit in 52c13 ways..then these can go to four people in 4 ways.hence prob is 4/52c13
@rubikmath said:you are mentioning ' in any order ' .but 12! is ordering .so your answer, divide it by 13! to take care of that ordering .
yeah it is right, either ways. I was actually making mistake in the ordering of the denominator. 
A tennis tournament has 8 players (A to H). There are 3 levels : Quarterfinals , semifinals and finals . Players are grouped randomly for the Quarterfinals . Every player has a probability of winning as 1/2 against any other player. What is the probability that A and B meet in a match during the tournament ?