What is the probability of spinning a wheel that has four colors (Red,Blue,Green and Yellow) and landing on the color red two times in a row?
[Math] the probability of spinning a roulette wheel that has four colors (Red,Blue,Green and Yellow) and landing on the color red two times in a row
probability
Related Solutions
Since Paul is presumably interested in eating the candies he selects, it is reasonable to assume that the candies are chosen without replacement. There is also an implicit assumption that the candies are selected without looking in the bag, which ensures that the selection is random.
You made the assumption that the selections are made successively, which is not necessary. We consider two approaches:
- The candies are selected randomly in succession without replacement.
- Two candies are selected randomly at once.
The candies are selected randomly in succession without replacement:
There are $6 + 5 + 12 + 9 + 13 = 45$ candies in the bag, of which $9$ are yellow and $45 - 9 = 36$ are not yellow. Thus, on the first draw, the probability that the first candy is not yellow is $36/45$. If the first candy is not yellow, that leaves $44$ candies in the bag, of which $35$ are not yellow. Thus, the probability that the second candy is also not yellow is $35/44$. Hence, the probability that both candies are not yellow is $$\frac{36}{45} \cdot \frac{35}{44} = \frac{4}{5} \cdot \frac{35}{44} = \frac{7}{11}$$
Two candies are selected randomly at once:
There are $\binom{45}{2}$ ways to select a subset of two of the $45$ candies and $\binom{36}{2}$ ways to select a subset of two of the $36$ candies that are not yellow. Hence, the probability that the two candies selected are not yellow is $$\frac{\dbinom{36}{2}}{\dbinom{45}{2}} = \frac{\dfrac{36!}{2!34!}}{\dfrac{45!}{2!43!}} = \frac{\dfrac{36 \cdot 35 \cdot 34!}{2 \cdot 1 \cdot 34!}}{\dfrac{45 \cdot 44 \cdot 43!}{2 \cdot 1 \cdot 43!}} = \frac{\dfrac{36 \cdot 35}{2}}{\dfrac{45 \cdot 44}{2}} = \frac{36 \cdot 35}{2} \cdot \frac{2}{45 \cdot 44} = \frac{36 \cdot 35}{45 \cdot 44} = \frac{7}{11}$$
Edit after rephrasing of the question. We have a row of $6$ flowers.
You want the probability that both the red and the blue flowers are on the same side of the green. One way to think of it is that all that matters is the relative position of these three flowers. There are $3!$ ways to order only the three of them, and in $2$ of these the green flower is in the middle, which is not the event we want. Thus, the probability is $4/6 = 2/3$.
A more detailed explanation
Suppose that you are considering all orderings of $6$ flowers. In your reasoning in the question you first pick places for red, blue and green and then permute the flowers, but not in a correct way.
We do something similar. Abbreviate each color by its initial. Suppose you chose positions $2, 3$ and $5$ for RBG, and positions $1$, $4$ and $6$ for YWP. One possible ordering in this case is $$ Y, R, B, W, G, P $$
Consider now that positions of $Y$, $W$ and $P$ are fixed and we are allowed to permute $R$, $B$ and $G$. Then, the possible colorings are $$ 1. \quad Y, R, B, W, G, P\\ 2. \quad Y, R, G, W, B, P\\ 3. \quad Y, B, R, W, G, P\\ 4. \quad Y, B, G, W, R, P\\ 5. \quad Y, G, B, W, R, P\\ 6. \quad Y, G, R, W, B, P $$ and from these $6$ orderings, only $4$ of them obey the rule of red and blue on the same side of green (or how I rephrased it, 'green not in the middle'). Finally, notice that we obtain all possible orderings by summing over all possible positions and fixing some order for $Y$, $W$ and $P$. Since for each choice of position and order of $WYP$ we have $4/6$ cases of 'green not in the middle', then we conclude that the probability of this event is $4/6$.
Best Answer
What is the probability you get red with a single roll. $\frac{1}{4}$
Suppose you allready had the luck of getting red (this happens $\frac{1}{4}$'th of the times.)
Then the odds you roll it again is $\frac{1}{4}$ (so this happens $\frac{1}{4}$'th of $\frac{1}{4}$'th of the times.)
So it happens $\frac{1}{16}$'th of the times. In other words the probability is $\frac{1}{16}$