# Using Draw Blanks approach to solve a probability question of drawing 2 red, 2 blue and 1 green marbles.

Q: "There are 6 children at a family reunion, 3 boys and 3 girls. They will be lined up single-file for a photo, alternating genders. How many arrangements of the children are possible for this photo?"

A: if we use the drawing 6 blanks approach _ _ _ _ _ _, we get (3*3*2*2*1*1) + (3*3*2*2*1*1) = 36 + 36 = 72.

so I want to try to use this same approach to answer this below question: "A bag holds 4 red marbles, 5 blue marbles, and 2 green marbles. If 5 marbles are selected one after another without replacement, what is the probability of drawing 2 red marbles, 2 blue marbles, and 1 green marble?"

my Answer: _ _ _ _ _ (4*3*5*4*2) divide by total probability which is 5! , which is wrong.

I know that we can easily answer this question using: $$\dfrac{(_4C_2) \cdot (_5C_2) \cdot (_2C_1)}{_{11}C_5}$$ I can also use permutations to answer this: $$\binom{5}{2}\binom{3}{2}\binom{1}{1}\left(\frac{4}{11}\right)\left(\frac{3}{10}\right)\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)\left(\frac{2}{7}\right) = \frac{20}{77}$$ But I am trying to learn how I can use different methods to solve the same problem.

Any help is greatly appreciated.

• What you wish to do is not advisable. There are $\binom{5}{2}\binom{3}{2}\binom{1}{1} = 10 \cdot 3 \cdot 1 = 30$ such sequences you would have to consider. As for the first problem, notice that we arrange the boys in $3!$ orders, arrange the girls in $3!$ orders, and decide whether a boy or girl is in the leftmost position in $2$ ways, giving $\binom{2}{1}3!3! = 72$ possible arrangements. – N. F. Taussig Apr 12 at 10:28
• true, makes perfect sense but if we were to do it using the non-advisable method. if possible, could u please write just one or 2 sequences out of the 30, just to help me see what those sequences would look like. would really help me understand better. Only If possible. – Lucky Apr 12 at 13:46
• Sure. The probability of obtaining blue, blue, red, red, green in that order is $\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$, while the probability of obtaining red, green, blue, blue, red in that order is $\frac{4}{11} \cdot \frac{2}{10} \cdot \frac{5}{9} \cdot \frac{4}{8} \cdot \frac{3}{7}$. – N. F. Taussig Apr 12 at 17:25
• If you compare these terms, you will notice that the denominators are the same and the numerators have just been permuted. Therefore, for each of the $30$ sequences, you will have a term equal to $\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$. Consequently, you can simply multiply $\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$ by $\binom{5}{2}\binom{3}{2}\binom{1}{1} = 30$ to get the correct probability. I hope that clarifies things somewhat. – N. F. Taussig Apr 12 at 17:27
• so if the question were "how many arrangements are possible of the 2 blue, 2 red and 1 green marbles" then the answer would be: 4*3*5*4*2*30? yea? – Lucky Apr 12 at 17:42

There are $$\binom{5}{2}$$ possible positions for the blue balls in the sequence. That leaves three positions in the sequence, of which two must be filled with red balls, so there are $$\binom{3}{2}$$ positions for the red balls in the sequence. The remaining position in the sequence must be filled with the green ball. Consequently, there are $$\binom{5}{2}\binom{3}{2}\binom{1}{1} = 10 \cdot 3 \cdot 1 = 30$$ possible sequences with two blue balls, two red balls, and one green ball.
The probability of obtaining blue, blue, red, red, green in that order is $$\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$ The probability of obtaining blue, green, red, red, blue in that order is $$\frac{5}{11} \cdot \frac{2}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$ Notice that the denominators are the same and that the terms in the numerator have just been permuted. Therefore, the probabilities of these sequences are equal. In fact, this is true of each of the $$30$$ sequences consisting of two blue balls, two red balls, and one green ball.
Therefore, while the probability of any particular sequence is $$\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$ the probability that a sequence will contain two blue balls, two red balls, and one green ball can be obtained by multiplying the probability that any particular sequence will appear by the $$\binom{5}{2}\binom{3}{2}\binom{1}{1} = 30$$ possible orders in which these balls could appear.
Thus, the probability of obtaining two blue balls, two red balls, and one green ball when five balls are selected from a bag containing five blue balls, four red balls, and two green balls is $$\binom{5}{2}\binom{3}{2}\binom{1}{1}\frac{5}{11} \cdot \frac{4}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}$$
Since we actually care about which balls are selected rather than the order in which they are selected, we do not need to consider the sequence in which the balls are selected. We just need to find the probability of selecting two of the five blue balls, two of the four red balls, and one of the two green balls when selecting five of the eleven balls in the bag, which is $$\frac{\dbinom{5}{2}\dbinom{4}{2}\dbinom{2}{1}}{\dbinom{11}{5}}$$