@krum said:7.5 min to nai ?
nai bhai.. Ijust loved this question when i saw the solution :P
@ScareCrow28 said:nai bhai.. Ijust loved this question when i saw the solution
subah subah bacchon ki lene par utar aae ap to
@ScareCrow28 said:A mouse has 3 rooms to go into. If it goes into room 1 it will find the cheese after 3 minutes. If it goes into room 2 it will look for cheese for 4 minutes, won't find it, and will go out. If it goes into room 3 it will look for cheese for 5 minutes, won't find it, and will go out.The mouse will not remember that it was in rooms 2 and 3 after it goes out of them, and it will continue going in and out until it finds the cheese. (It can go into the same room again and again.)What is the average time for the mouse to find cheese?
Let the average time to find cheese be T. The mouse will have three choices each chosen with 1/3 probability:
1. use 3 minutes to find the cheese,
2. use 4 minutes and try again (4 + T), or
3. use 5 minutes and try again (5 +T).
The equation would be: T = 1/3 * ((3) + (4 + T) + (5 + T))
Hence from here you get : T = 12 minutes
2. use 4 minutes and try again (4 + T), or
3. use 5 minutes and try again (5 +T).
The equation would be: T = 1/3 * ((3) + (4 + T) + (5 + T))
Hence from here you get : T = 12 minutes
@krum @ayushnasa @culdip Sir..
P.S. Some doubts will always hover around the solution..but this is what the solution says..
Off to lunch
@ScareCrow28 said:A mouse has 3 rooms to go into. If it goes into room 1 it will find the cheese after 3 minutes. If it goes into room 2 it will look for cheese for 4 minutes, won't find it, and will go out. If it goes into room 3 it will look for cheese for 5 minutes, won't find it, and will go out.The mouse will not remember that it was in rooms 2 and 3 after it goes out of them, and it will continue going in and out until it finds the cheese. (It can go into the same room again and again.)What is the average time for the mouse to find cheese?
@ScareCrow28 said:Let the average time to find cheese be T. The mouse will have three choices each chosen with 1/3 probability:
1. use 3 minutes to find the cheese,
2. use 4 minutes and try again (4 + T), or
3. use 5 minutes and try again (5 +T).
The equation would be: T = 1/3 * ((3) + (4 + T) + (5 + T))
Hence from here you get : T = 12 minutes
o.O WHAT?!!
The solution is sooo smooth!!
Three Professors Dr. Gupta, Dr. Sharma and Dr. Singh are evaluating answer scripts of a subject. Dr. Gupta is 40% more efficient than Dr. Sharma, who is 20% more efficient than Dr. Singh. Dr. Gupta takes 10 days less than Dr. Sharma to complete the evaluation work. Dr. Gupta starts the evaluation work and works for 10 days and then Dr. Sharma takes over. Dr. Sharma evaluates for next 15 days and then stops. In how many days, Dr. Singh can complete the remaining evaluation work.
A. 7.2 days
B. 9.5 days
C. 11.5 days
D. None of these
@surajsrivastav said:Three Professors Dr. Gupta, Dr. Sharma and Dr. Singh are evaluating answer scripts of a subject. Dr. Gupta is 40% more efficient than Dr. Sharma, who is 20% more efficient than Dr. Singh. Dr. Gupta takes 10 days less than Dr. Sharma to complete the evaluation work. Dr. Gupta starts the evaluation work and works for 10 days and then Dr. Sharma takes over. Dr. Sharma evaluates for next 15 days and then stops. In how many days, Dr. Singh can complete the remaining evaluation work.A. 7.2 days B. 9.5 days C. 11.5 days D. None of these
Sorrry 😛 A..7.2 days
@surajsrivastav said:Three Professors Dr. Gupta, Dr. Sharma and Dr. Singh are evaluating answer scripts of a subject. Dr. Gupta is 40% more efficient than Dr. Sharma, who is 20% more efficient than Dr. Singh. Dr. Gupta takes 10 days less than Dr. Sharma to complete the evaluation work. Dr. Gupta starts the evaluation work and works for 10 days and then Dr. Sharma takes over. Dr. Sharma evaluates for next 15 days and then stops. In how many days, Dr. Singh can complete the remaining evaluation work.A. 7.2 days B. 9.5 days C. 11.5 days D. None of these
Yup! 7.2
@surajsrivastav said:Three Professors Dr. Gupta, Dr. Sharma and Dr. Singh are evaluating answer scripts of a subject. Dr. Gupta is 40% more efficient than Dr. Sharma, who is 20% more efficient than Dr. Singh. Dr. Gupta takes 10 days less than Dr. Sharma to complete the evaluation work. Dr. Gupta starts the evaluation work and works for 10 days and then Dr. Sharma takes over. Dr. Sharma evaluates for next 15 days and then stops. In how many days, Dr. Singh can complete the remaining evaluation work.A. 7.2 days B. 9.5 days C. 11.5 days D. None of these
iska 7.2 hai, 3-4 din pehle hi kiya tha
@surajsrivastav said:Three Professors Dr. Gupta, Dr. Sharma and Dr. Singh are evaluating answer scripts of a subject. Dr. Gupta is 40% more efficient than Dr. Sharma, who is 20% more efficient than Dr. Singh. Dr. Gupta takes 10 days less than Dr. Sharma to complete the evaluation work. Dr. Gupta starts the evaluation work and works for 10 days and then Dr. Sharma takes over. Dr. Sharma evaluates for next 15 days and then stops. In how many days, Dr. Singh can complete the remaining evaluation work.A. 7.2 days B. 9.5 days C. 11.5 days D. None of these
A.... 7.2 Days
assume Dr.singh can do 100x units of work per day
so Dr. Sharma can do 120x units
and Dr. gupta can do 168x units
given 120x*T = 168x*(T-10)
solve u get T = 35... so total work is 4200x units
Gupta can do 1680 units of work in 10 days
Sharma can do 1800 units of work in 15 days
total = 1680+1800 = 3480
remaiining = 720 ..
so 7.2 Days
@ayushnasa said:@ScareCrow28 Aapke probability ke sawal khatam? -_-
aur kuch bhi post karne do usko, varna mein depression mein chala jaunga :P
@ayushnasa said:@ScareCrow28 Aapke probability ke sawal khatam? -_-
More probablity?? Abi dalta hun :P
@surajsrivastav said:@ayushnasa@ScareCrow28 Approch to btao???OA is A i.e. 7.2
Efficiancies:
Singh=1
Sharma=1.2
Gupta=1.2*1.4
Let z be the amount of work:
Then,
{z/(1.4*1.2)}+10 = z/1.2
Solving z=42..
Now,
Gupta in 10 days=1.4*1.2*10=16.8
Sharma in 15 days = 1.2*15 =18
Remaining work=42-16.8-12=7.2
Time taken by singh=7.2/1=7.2`
A queue of n + m people is waiting at a box office; n of them have 5-pound notes and m have 10-pound notes. The tickets cost 5 pounds each. When the box office opens there is no money in the till. If each customer buys just one ticket, what is the probability that none of them will have to wait for change?
@ScareCrow28 said:A queue of n + m people is waiting at a box office; n of them have 5-pound notes and m have 10-pound notes. The tickets cost 5 pounds each. When the box office opens there is no money in the till. If each customer buys just one ticket, what is the probability that none of them will have to wait for change?
: n>=m right??
@ScareCrow28
@ScareCrow28 said:A queue of n + m people is waiting at a box office; n of them have 5-pound notes and m have 10-pound notes. The tickets cost 5 pounds each. When the box office opens there is no money in the till. If each customer buys just one ticket, what is the probability that none of them will have to wait for change?
[(n!*m!) + ((n+1) P m * n!) ]/(n+m)!
@ScareCrow28 said:A queue of n + m people is waiting at a box office; n of them have 5-pound notes and m have 10-pound notes. The tickets cost 5 pounds each. When the box office opens there is no money in the till. If each customer buys just one ticket, what is the probability that none of them will have to wait for change?
(n-m)/(n+m) ?
@Brooklyn said:: n>=m right??
Yes.. n>=m 😃 nai to question galt hojaega..aur tum log mere upar chad jaoge :P