A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement.

What is the probability that the first two candies drawn are orange and the third is red?

To find the probability of drawing two orange candies first, followed by a red candy, we will calculate the probability of each event occurring in sequence, considering that there is no replacement. This means the number of candies in the bag decreases as each candy is drawn.

### Step 1: Total number of candies
The total number of candies in the bag is:

\[
12 \, (\text{red}) + 12 \, (\text{blue}) + 7 \, (\text{green}) + 13 \, (\text{brown}) + 3 \, (\text{orange}) + 10 \, (\text{yellow}) = 57 \, (\text{total candies})
\]

### Step 2: Probability of drawing the first orange candy
There are 3 orange candies out of the 57 total candies, so the probability of drawing an orange candy on the first draw is:

\[
P(\text{first orange}) = \frac{3}{57}
\]

### Step 3: Probability of drawing the second orange candy
After drawing one orange candy, there are now 2 orange candies left and a total of 56 candies remaining. Therefore, the probability of drawing another orange candy on the second draw is:

\[
P(\text{second orange}) = \frac{2}{56}
\]

### Step 4: Probability of drawing a red candy
After two orange candies have been drawn, there are 12 red candies remaining out of the 55 total remaining candies. Thus, the probability of drawing a red candy on the third draw is:

\[
P(\text{third red}) = \frac{12}{55}
\]

### Step 5: Calculate the overall probability
The overall probability is the product of these three probabilities:

\[
P(\text{first two orange, third red}) = \frac{3}{57} \times \frac{2}{56} \times \frac{12}{55}
\]

Now, multiply the fractions:

\[
P = \frac{3 \times 2 \times 12}{57 \times 56 \times 55} = \frac{72}{175560}
\]

Simplifying the fraction:

\[
P = \frac{1}{2430}
\]

Thus, the probability of drawing two orange candies followed by one red candy is:

\[
\boxed{\frac{1}{2430}}
\]

Would you like more details or have any questions about this? 

Here are 5 related questions:
1. What is the probability of drawing three red candies in a row?
2. How would the probability change if the candies were replaced after each draw?
3. What is the probability of drawing no orange candies in the first three draws?
4. How many ways can you draw two orange candies followed by one red candy in this situation?
5. If four candies were drawn, what is the probability that two are orange and two are red?

**Tip**: When dealing with probabilities without replacement, remember to adjust the total number of items after each event.

A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement. What is the probability that the first two candies drawn are orange and the third is red?

Learn how to calculate the probability of drawing two orange candies followed by one red candy from a bag of M&Ms, using the concept of probability without replacement. This step-by-step solution covers the multiplication rule and probability concepts for Grades 9-11.

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