ShoutBox (Part 1)

Jobs & Careers
16214
@abhimukh19 said:
@brixcel i think we broke the rhythm of some studious people on SBT ,suddenly quant question discussion has stopped
hehehee 😁 :splat: that keeps happening... People are used to it 😁 :mg:
16215
@ankita14 said:
Humka bhi shaadi karni hai :/
kauno nalayak ko pasand kiye ka abhi ? 😁 😁
16216

anyways if anybody is interested in discussing marketing , finance , HR , IT , economics .. etc m always up for it :mg:

16217
@sameersapre23 said:
A die is thrown and a coin is then tossed as many times as the number shown by the die. What is the probability of getting more number of heads than tails?
(47 ,87,77,97) / 192
Probability of each number on die = 1/6

E = Event when head occurs more number of times than tails

When 1 is rolled: P(E) = 1/2 (H)
When 2 is rolled: P(E) = 1/4 (HH)
When 3 is rolled: P(E) = 1/8 (HHH) + 3/8 (HHT)
When 4 is rolled: P(E) = 1/16 (HHHH) + 4/16 (HHHT)
When 5 is rolled: P(E) = 1/32 (HHHHH) + 5/32 (HHHHT) + 10/32 (HHHTT)
When 6 is rolled: P(E) = 1/64 (HHHHHH) + 6/64 (HHHHHT) + 15/64 (HHHHTT)

1/6 *( 1/2 + 1/4 + 1/2 + 5/16 + 1/2 + 22/64) = 77/192

Alternative:

Whenever a die is thrown odd number of times, there are equal cases when Heads are more than Tails, and Tails are more than heads, -> P(H more) = 1/2

Whenever a die is thrown even number of times, there are three types of cases, Heads = Tails, Heads > Tails , Tails > Heads. If we remove the Heads = Tails case, and divide the remaining probability by 2, we will get P(H more) = 1/2 * ( 1 - P(H=T))

Using the above we get 1/6 *( 1/2 + 1/4 + 1/2 + 5/16+ 1/2 + 22/64) = 77/192..

Second method is shorter I think...

PS: Figured out the shorter method while doing the question.. Damn !
16218
Two white squares and a black square have to be chosen from a chess board such that no two of them are in the same row or column. In how many ways can the selection be made?

6400
2400
3200
none of these........!!
16219
@abhimukh19 said:
M sure gangs of wasseypur ka hi koi hoga
No way , shadi se vishwaash uth gya hai mera
Gangs of Wasseypur...maine nahi dekhi abhi :|
16220
@YouMadFellow said:
Chalega .. Jaate jaate wo geometry ke question ka solution bata do ! Square me maximum R of semi-circle...
Sirjee.. answer is 1) option ... [2 - root(2)] hai!
16221
@ankita14 said:
Humka bhi shaadi karni hai :/
o.O
16222
@abhimukh19 said:
anyways if anybody is interested in discussing marketing , finance , HR , IT , economics .. etc m always up for it
16223
@hanushanand said:
Sirjee.. answer is 1) option ... [2 - root(2)] hai!
Maine solution maanga tha .. Dusre wale ka bhi de do.. x^2 + 1/x^2 ... Geometry weak hai :splat:
16224
@ankita14 said:
Humka bhi shaadi karni hai :/
Go ahead :mg: next month karna ,Catering ,Tent pe discount milega season araha hai :p
16225
@brixcel said:
kauno nalayak ko pasand kiye ka abhi ?
Kono nalayak humka pasanda hi naahi karat :splat:
16226
@abhimukh19 said:
Go ahead next month karna ,Catering ,Tent pe discount milega season araha hai
Kono ladka jaanat ho? :lookround:
16227
@Ibanez said:
o.O
Ka? :/
16228
@YouMadFellow said:
Probability of each number on die = 1/6
E = Event when head occurs more number of times than tails
When 1 is rolled: P(E) = 1/2 (H)
When 2 is rolled: P(E) = 1/4 (HH)
When 3 is rolled: P(E) = 1/8 (HHH) + 3/8 (HHT)
When 4 is rolled: P(E) = 1/16 (HHHH) + 4/16 (HHHT)
When 5 is rolled: P(E) = 1/32 (HHHHH) + 5/32 (HHHHT) + 10/32 (HHHTT)
When 6 is rolled: P(E) = 1/64 (HHHHHH) + 6/64 (HHHHHT) + 15/64 (HHHHTT)
1/6 *( 1/2 + 1/4 + 1/2 + 5/16 + 1/2 + 22/64) = 77/192
Alternative:
Whenever a die is thrown odd number of times, there are equal cases when Heads are more than Tails, and Tails are more than heads, -> P(H more) = 1/2
Whenever a die is thrown even number of times, there are three types of cases, Heads = Tails, Heads > Tails , Tails > Heads. If we remove the Heads = Tails case, and divide the remaining probability by 2, we will get P(H more) = 1/2 * ( 1 - P(H=T))
Using the above we get 1/6 *( 1/2 + 1/4 + 1/2 + 5/16+ 1/2 + 22/64) = 77/192..
Second method is shorter I think...
PS: Figured out the shorter method while doing the question.. Damn !
gud one .........sir....!!yes it is shorter.........!! but waha cat mai click hona chayee....!!
second method mujhe kabhi click ni hota.........!!
16229
@ankita14 said:
Kono ladka jaanat ho?
Haan janta to main Salman Khan ko bhi hu.. but wo muje nai janta :mg:

@naga25french sir have u seen the documentary on enron fraud?
16230
@ankita14
@ankita14 said:
Kono ladka jaanat ho?
dehati bhasha ka prayog
16231
@ankita14 said:
Kono nalayak humka pasanda hi naahi karat
ee to gambhir samasya khadi hoi gayi....photu-wa mein to tum achchhi dikhat ho...fir kaahe aisan bolat ho ? :mg:
16232
@YouMadFellow said:
Probability of each number on die = 1/6
E = Event when head occurs more number of times than tails
When 1 is rolled: P(E) = 1/2 (H)
When 2 is rolled: P(E) = 1/4 (HH)
When 3 is rolled: P(E) = 1/8 (HHH) + 3/8 (HHT)
When 4 is rolled: P(E) = 1/16 (HHHH) + 4/16 (HHHT)
When 5 is rolled: P(E) = 1/32 (HHHHH) + 5/32 (HHHHT) + 10/32 (HHHTT)
When 6 is rolled: P(E) = 1/64 (HHHHHH) + 6/64 (HHHHHT) + 15/64 (HHHHTT)
1/6 *( 1/2 + 1/4 + 1/2 + 5/16 + 1/2 + 22/64) = 77/192
Alternative:
Whenever a die is thrown odd number of times, there are equal cases when Heads are more than Tails, and Tails are more than heads, -> P(H more) = 1/2
Whenever a die is thrown even number of times, there are three types of cases, Heads = Tails, Heads > Tails , Tails > Heads. If we remove the Heads = Tails case, and divide the remaining probability by 2, we will get P(H more) = 1/2 * ( 1 - P(H=T))
Using the above we get 1/6 *( 1/2 + 1/4 + 1/2 + 5/16+ 1/2 + 22/64) = 77/192..
Second method is shorter I think...
PS: Figured out the shorter method while doing the question.. Damn !
Ye sab aapne type kiya kya??
16233
@ankita14 said:
Ka? :/
Invite zaroor karna.