probability & Perm Qs (1 Viewer)

[M@C D@DDY]

I just need some help with these two qs fanx!!

Rwo people john and Kim play a dice game. john throws two dice and kim throws one kim wins if john fails to beat his number with either one of his two dice(clearly,if all the numbers are the same kim wins) wat is the probablility that kim wins the game?

dunno if qs is printed wrong but i don't get it.....

containers are coded by different arrangements of couloured dots in a row. the colours r red white and blue.

In an arrangement at most 3 of the dots are red at most 2 r white and at most one is blue

Find the number of different codes possible if 6 dots r used

on some containers onli five dots are used. find the number of different codes possible in this case. jusitify your answer

thses are from skool past papers i srsli have no clue any help would b appreciated fanx!

 

[ND]

Originally posted by M@C D@DDY

Rwo people john and Kim play a dice game. john throws two dice and kim throws one kim wins if john fails to beat her number with either one of his two dice(clearly,if all the numbers are the same kim wins) wat is the probablility that kim wins the game?

I think that's what it's supposed to say. I.e. kim throws one dice, then for john to win, he must beat kim's dice, and he gets two throws.

If kim throws a 1, his probability of winning is (5/6)^2, if she throws a 2 it's (4/6)^2 etc. Sum these up then divide by 6 (cos each outcome is equally likely) to get the prob of john winning.

For the 2nd q, the prob is just 6!/3!*2! cos there are 6 objects, and 3 and 2 of them are the same. The the 2nd part, take cases for when the blue is excluded, when the white is excluded, and when teh red is excluded.

 

[M@C D@DDY]

but it says his not HERS dat's the problem

 

[Xayma]

Originally posted by ND
If kim throws a 1, his probability of winning is (5/6)^2, if she throws a 2 it's (4/6)^2 etc. Sum these up then divide by 6 (cos each outcome is equally likely) to get the prob of john winning.

If kim throws a 1 the chances of her winning is 1/36 isnt it (since John must throw a 1,1) therefore Johns chances are 35/36

If kim throws a 2, the probablilty of her winning is (2/6)²

If kim throws a 3 the probability of her winnign is (3/6)² etc.

Therefore the probability is (1²+2²+3²+4²+5²+6²)/6³

They way you did it is if you are doing if he gets a 1 he loses.

 

[ND]

I was doing john's probability of winning, not kim's (and i made the exact same typo that was in the original question). To get to kim's prob of winning from john's, do 1 minus it.

 

[Xayma]

I know that but you were still incorrect Johns chances of winning if kim throws a 1 is 35/36 the way you were doing it is if he throws a 1 he loses.

Its easy to tell you have gone wrong as 5/6>(5/6)²

 

[CM_Tutor]

John's probability of winning if Kim throws a 1 is not (5 / 6)², as you wrote. It is 1 - (1 / 6)² = 35 / 36, as Xayma pointed out. The probability of John winning if Kim threw a 1 would only be (5 / 6)² if John had to beat 1 with both his dice, but he doesn't. He only has to beat it with at least 1 of his dice.

 

[ND]

Hahah yeh i had that in my head but somehow (5/6)^2 got onto the paper (well, onto the screen) and from there... that's the 2nd time i've done something really really stupid today... i don't know what's getting into me... =/ anyway that's enough for me today, off to see kill bill!

 

[Archman]

for q2 second part: the answer is the same as the first part, since exactly one dot is not used, for every coding from the first part, we can just remove the last dot and get a unique arrangement.