Y4ss

[Y4ss @Y4ss12]

@quantymacro Deadlift is not mandatory if you just want to be healthy/good looking for summer. Plenty of safer exercise for posterior chain

Should also work your core, if you are staying in seated position for 7hrs+/day your core is probably a weak link

Stay safe !

[Y4ss @Y4ss12]

@quantymacro You would expect some soreness but it has to be muscular soreness. Be careful out there deadlift can be a dangerous lift, especially if you have long femur and short torso.

[Y4ss @Y4ss12]

@quant_prep Ask each children to choose H or T and flip the coin and then you keep winner(s) and repeat (they must not choose the same for the 2nd stage)

If you have only 1 winner at stage one, he must choose again and if he loss you start all over.

In any case they have same prob of win

[Y4ss @Y4ss12]

@quant_prep Full solution algo implementation below :

https://t.co/C0giisL5QK

[Y4ss @Y4ss12]

@quant_prep 0.8 x 0.6 gives 0.48 but you have to assume independence of both events.

Good follow up questions what is the probability of rain on Sunday given rain on Saturday. You would need to know the probability of rain on Saturday given rain on Sunday and use bayes theorem

[Y4ss @Y4ss12]

@charafzgu Also the your travel stations would also not be all equally likely and would probably follow somekind of pattern (like the one close to your home would come more often than the one that are very far)

[Y4ss @Y4ss12]

@charafzgu Good probability problem but in reality it would be even more complex

A control occuring at the station you are located doesn't necessarly mean that you are getting caught (could also be considered a chance process) and all stations are not equally likely of being controlled.

[Y4ss @Y4ss12]

@SylvainReloaded @quant_prep As promised the full algo implementation in Python, enjoy !

https://t.co/1Ibqt0AuTP

[Y4ss @Y4ss12]

@SylvainReloaded @quant_prep Dont' know if my explanation is clear 😅

If i have time this weekend i'll upload the algorithm implementation this week-end

[Y4ss @Y4ss12]

@iamalabradoodle @SylvainReloaded @quant_prep If 1st measurement is different Same logic, you replace one with an unmeasured and you can isolate the odd one

Anyway, if you can crack this on the fly during an interview without knowing solution beforehand you are somekind of genius

[Y4ss @Y4ss12]

@iamalabradoodle @SylvainReloaded @quant_prep It does works but i agree too long for Twitter.

Outcome A is the easiest, 4 candidate and two measurement remaining

If 1st measurement is identic then you replace one with an unmeasured, if identic again it's the remaining one if not it's the one you measured in last.

[Y4ss @Y4ss12]

@quant_prep Outcome B

Weight are different on the first measurement so we cannot isolate the group of 4 but we have a group of 8.

We add one ball from group of 4 to group of 8 to make it group of 9

And then proceed to isolate the odd ball by splitting group of 9 into 3 group of 3

[Y4ss @Y4ss12]

@quant_prep Outcome A)

The remaining group of 4 contain the odd ball and we have two measurement left.

Pick two ball, measure them

If they are equal, replace one of them with an unmeasured one and re-measure. This way you isolats the odd one

Same procedure if weight are different