probability problem (1 Viewer)

[jake2.0]

in this question; 3 six-sided dice are thrown once, what is the probablity that exactly 2 of the dice show 2.

y can't u just say that the probability = 1/6.1/6.5/6? but instead must use binomial probabilty and say 3C2.1/6.1/6.5/6. whats the point of having the 3C2 there??

thanks guys

 

[acmilan]

3C2 = 3

There are 3 possible combinations:

(2, 2, number other than 2)
(2, number other than 2, 2)
(number other than 2, 2, 2)

Hence you need the 3C2

Doing it the 2 unit way, the answer is:

(1/6.1/6.5/6) + (1/6.5/6.1/6) + (5/6.1/6.1/6)

But this just equals 3(1/6.1/6.5/6), ie. 3C2(1/6.1/6.5/6)

 

[jake2.0]

but i thought that order wasn't important as it is not a permutation

 

[香港!]

in this question; 3 six-sided dice are thrown once, what is the probablity that exactly 2 of the dice show 2.

y can't u just say that the probability = 1/6.1/6.5/6? but instead must use binomial probabilty and say 3C2.1/6.1/6.5/6. whats the point of having the 3C2 there??

thanks guys

Helloooo!
My probability is horrible... so just practicing here... don't laugh at me if i'm wrong k?

"3 six-sided dice are thrown once, what is the probablity that exactly 2 of the dice show 2. "

P(get a 2 on one of the dice)=1\6
so if 2 of the dices show a 2, then it's (1\6)²
the remainder must NOT show this 1\6, so it's 5\6
This gives:
P(exactly 2 of the dice show 2)= 3C2 (5\6)(1\6)²
the 3C2 is there to account for the 2 dices showing 2 and the other dice, they can be arranged in diff. ways as well....

Correct me if i'm wrong

 

[jake2.0]

they can be arranged in diff. ways as well....

but i thought that order wasn't important as it is not a permutation

 

[Jago]

(2, 2, number other than 2) does NOT equal (2, number other than 2, 2) which also does NOT equal (number other than 2, 2, 2)

hence the 3C2