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KEITA FC 8C
11/12/2017 at 21:34KEITA FC 8C
11/12/2017 at 21:48dunglamtoan
13/12/2017 at 08:24Kaya Renger Coodinator
13/12/2017 at 13:05
It is impossible to combine only 2 differentvalue tokens because 5+19, 11+19, 5+11 \(⋮̸\) 56
Add three different tokens we have: \(5+11+19=35\left(points\right)\)
We have: \(5635=21\) (points) left when we add two tokens.
Because 21 points worth 2 5point tokens and a 11point token so in the box, there are:
+ 3 5point tokens
+ 2 11point tokens
+ 1 19point tokens

FA Liên Quân Garena 30/12/2017 at 21:48
it is impossible to combine only 2 differentvalue tokens because 5+19, 11+19, 5+11 ⋮̸
56
Add three different tokens we have: 5+11+19=35(points)
We have: 56−35=21
(points) left when we add two tokens.
Because 21 points worth 2 5point tokens and a 11point token so in the box, there are:
+ 3 5point tokens
+ 2 11point tokens
+ 1 19point tokens

Whatever x is, the median will still be 5 or 6 (If x > 5 than the median is 5 and 6, if x < 5 than the median is 5 and if x = 5 than the median will be 5). So there will be 2 possible values for x.

FA Liên Quân Garena 30/12/2017 at 21:49
Whatever x is, the median will still be 5 or 6 (If x > 5 than the median is 5 and 6, if x < 5 than the median is 5 and if x = 5 than the median will be 5). So there will be 2 possible values for x.

The probability that Xera has a four is: \(1:6=\dfrac{1}{6}\)
The probability that Yeta has a four is: \(4:52=\dfrac{1}{13}\) (A standard deck has 4 cards of four)
So the probability that Yeta wins is: \(\dfrac{1}{13}:\left(\dfrac{1}{6}+\dfrac{1}{13}\right)=\dfrac{6}{19}\)

FA Liên Quân Garena 30/12/2017 at 21:49
The probability that Xera has a four is: 1:6=16
The probability that Yeta has a four is: 4:52=113
(A standard deck has 4 cards of four)
So the probability that Yeta wins is: 113:(16+113)=619