Here is a set of probability questions. I am having trouble with the last three.
Assume that you have three six-sided dice. One is yellow (Y), one is red (R) and one is white (W).
A)What is the probability that, in rolling only the red die, it will show the number 5?
B)What is the probability that, in rolling only the white die, it will show an even number?
C)What is the probability that, in rolling all three dice at once, the outcome will be 2(Y), 4(R), 6(W)?
D)What is the probability that, in rolling all three dice at once, the outcome will be 1, 3 and 5 in any color combination?
E)What is the probability that, in rolling all three dice at once, the outcome will be the same number on all three dice?
F)What is the probability that, in rolling all three dice at once, the outcome will be a different number on all three dice?
Here is what I have so far:
D) From part C I found that P for any # is $1/6$ thus $P(2Y,4R,6W)=(1/6)^3=1/216$. Therefore, same logic should apply here with the exception that there will be a lot more combinations w/o restriction on color. Since we have $3$ color boxes there would be total of $9$ combinations possible and thus $P=(1/6)^9$ but its not the correct answer. On the contrary, the right answer is $P=6(1/216)=1/36$ I DO NOT UNDERSTAND why we multiply by $6$.
E) here since they ask for the same number the probability of getting say $(2,2,2)$ combo is $P= 1/6 \cdot 1/6 \cdot 1/6 = 1/216$ thus having $6$ #'s on the dice the total probability of having all the same #'s is $P = 1/216+…+1/216=6/216=1/36$
F) I dont understand as well, since we found in E that having the same #'s on all three dice is $1/36$ then would not $P(\text{not the same #'s})= 1-1/36=35/36$? but for some reason it's not the correct answer
I would appreciate if you could explain your reasoning, I'm new to probability and pretty much forced to learn it for genetics problems.
Best Answer
D)
In the order 1,3,5, $Pr = \dfrac{1}{6}\cdot\dfrac{1}{6}\cdot\dfrac{1}{6} = \dfrac{1}{216}$
but there could be $3! = 6$ different orders, so multiply by 6
E)
First number can be anything, each of the next two have to be different, hence $\dfrac{1}{6}\cdot\dfrac{1}{6} = \dfrac{1}{36}$
F)
6 ways to get first number, 5 ways for second one, and 4 ways for third,
hence $\dfrac{6*5*4}{216} = \dfrac{5}{9}$
Note that the complement of all same is not all different, it is all not same