anonymous
  • anonymous
Fool's problem of the day, Four identical dice are tossed simultaneously. What is the probability that at least three of the four nos. shown are different?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
1-(1/6)(1/6)(5/6)(4/6)
experimentX
  • experimentX
4*(6*5*4)/6^4 + 6*5*4*3/6^4
anonymous
  • anonymous
@CoCoTsoi: Doesn't seem right.

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anonymous
  • anonymous
oops I will think again :D
Callisto
  • Callisto
6x5x4x3/ 6^4 + 6x5x4x3/6^4 =5/9 @callisto doesn't seem right :S
anonymous
  • anonymous
\( \frac{35}{54}\) is not the right answer. @experimentX
experimentX
  • experimentX
is it (6*5*4)/6^3 + 6*5*4*3/6^4
anonymous
  • anonymous
Yes, Experimentx Congratz :D Now please post the explanation/solution :)
experimentX
  • experimentX
hahah .. sometimes you are lucky. :D
Callisto
  • Callisto
Nah , I was right if i had not changed my mind ?! (-sorry - neglect it-)
anonymous
  • anonymous
The harder I work, the luckier I get ~Samuel Goldwyn.
experimentX
  • experimentX
well, ... i guess i have to agree. thank you for the problem. i am looking forward for others.
Callisto
  • Callisto
The harder i work , the more mistakes i make ~ Me :P Thanks too !!
anonymous
  • anonymous
I don't get the second part :-S
experimentX
  • experimentX
at least implies ... you have to consider the probability of 4 different also
anonymous
  • anonymous
the fourth dice becomes irrelevant, no?
anonymous
  • anonymous
The second part is self evident, in the first part he is considering 6^3 instead of 6^4. I think that's what he is referring as luck :P
experimentX
  • experimentX
yup ... you saw through it.
Callisto
  • Callisto
i don't understand the first and the second part I thought the first part 6x5x4x6 / 6^4 for the last one can be any numbers the second part, 6x5x4x3 / 6^4 for the last number can only be the number that has been 'selected' Is it like that?
anonymous
  • anonymous
6 / 6 cancel each other out, you could rewrite it 5*4/6^2. But I'm still confused about the second part
experimentX
  • experimentX
" .. probability that AT LEAST three of the four nos .." .. you also have to add the probability of getting 4 different nos
Callisto
  • Callisto
Ah, i got it reversed :S, thanks experimentX :)
anonymous
  • anonymous
ok, so, the important part is simply that 3 numbers are different, since the 4th can be whatever? so that would mean simply 6*5*4/6^3? (God I hate stats.)
anonymous
  • anonymous
No, that's not right, it's giving the right answer but it's the correct approach.
anonymous
  • anonymous
HINT: For the part exactly 3 different: \[ \frac 12 \binom{4}{3} \left( 6 \times 5 \times 4 \times 3 \right) \]
anonymous
  • anonymous
I would like to rephrase/edit "correct" to "clear"
anonymous
  • anonymous
ohhhh, ok, my piece of paper seems to be telling me something. I'm getting there lol
Callisto
  • Callisto
Sorry, I think I might be the one who still don't understand now :S
anonymous
  • anonymous
turns out I got the same result again... ok, if all 4 are different, the probability is 6*5*4*3/6^4. for exactly 3 different, it's 6*5*4*3/6^4. if you shave everything off of that, I still am getting 5*4/6^2
anonymous
  • anonymous
if you don't take into account the 4th dice, you get 6*5*4/6^3, but you can't then add 6*5*4*3/6^4, you're counting stuff twice
experimentX
  • experimentX
now i'm lost too
anonymous
  • anonymous
Ask Zarkon, he might help :)
experimentX
  • experimentX
I think @m_charron2 might be right ... the probability of getting exactly 3 different is 6*5*4*3/6^4 and of exactly 4 different is 6*5*4*3/6^4
experimentX
  • experimentX
adding up gives again 6*5*4/6^3
anonymous
  • anonymous
Meh, nah, I think I get it. I understand that, to get 3 different ones, you need 6*5*4/6^3, and that is because your fourth dice is hidden in there as 6/6 which you don't put. the fourth dice is already accounted for in the first part of the equation, because the 3/6 where it's different and the 3/6 where it's the same. Looks like a conceptual error. Either you don't check the 4th dice, which gives you 6*5*4/6^3 or you check it and you get 2(6*5*4*3/6^4)
anonymous
  • anonymous
Now, I'm getting myself a big coffee. That obscure part of my brain was all rusty.
experimentX
  • experimentX
for 4 rolls i guess
Callisto
  • Callisto
But then it's different from the answer given by FoolForMath :(
anonymous
  • anonymous
Okay, here are some more hints, if we consider two dices then (6,5) and (5,6) are considered two cases. but (6,6) and (6,6 are considered to be the same.
anonymous
  • anonymous
oh... I,m completely boggled in trying to understand it, but yeah, you were right... is it something like : (6*6*5*4+6*5*6*4+6*5*4*6+6*5*4*3)/6^4?
anonymous
  • anonymous
4*(6*5*4)/6^4 + 6*5*4*3/6^4
anonymous
  • anonymous
Case where 4 are different: 6 for first die, *5 for second, only 5 different options left *4 for third, same logic as above *3 for fourth Number of situations where 4 die rolled all have different readings. Dividing by the number of possible outcomes, 6^4 (6 for each die), it becomes 6*5*4*3/6^4 = 5*4*3/6^3 = 60/216 = 5/18 Case where 3 are different: 6 for first die, *5 for second, *4 for third, *6 for fourth because it doesn't matter what it is. Same as above, simplifies to 20/36 = 5/9 5/9 + 5/18 = 5/6

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