0.12 Probability  (Page 5/8)

Let $R$ be the event that the marble drawn is red, and let $W$ be the event that the marble drawn is white.

We draw the following tree diagram.

Although the tree diagrams give us better insight into a problem, they are not practical for problems where more than two or three things are chosen. In such cases, we use the concept of combinations that we learned in [link] . This method is best suited for problems where the order in which the objects are chosen is not important, and the objects are chosen without replacement.

Let us suppose the marbles are labeled as $R_1$ , $R_2$ , $R_3$ , $W_1$ , $W_2$ , $B_1$ , $B_2$ , $B_3$ .

1. $P\left(\text{Two red and one white}\right)$

   We analyze the problem in the following manner.

   Since we are choosing 3 marbles from a total of 8, there are $\mathrm{8C3}=\text{56}$ possible combinations. Of these 56 combinations, there are $\mathrm{3C2}\times \mathrm{2C1}=6$ combinations consisting of 2 red and one white. Therefore,

   $P\left(\text{Two red and one white}\right)=\frac{\mathrm{3C2}\times \mathrm{2C1}}{\mathrm{8C3}}=\frac{6}{\text{56}}$ .

2. $P\left(\text{One of each color}\right)$

   Again, there are $\mathrm{8C3}=\text{56}$ possible combinations. Of these 56 combinations, there are $\mathrm{3C1}\times \mathrm{2C1}\times \mathrm{3C1}=\text{18}$ combinations consisting of one red, one white, and one blue. Therefore,

   $P\left(\text{One of each color}\right)=\frac{\mathrm{3C1}\times \mathrm{2C1}\times \mathrm{3C1}}{\mathrm{8C3}}=\frac{\text{18}}{\text{56}}$ .

3. $P\left(\text{None blue}\right)$

   There are 5 non-blue marbles, therefore

   $P\left(\text{None blue}\right)=\frac{\mathrm{5C3}}{\mathrm{8C3}}=\frac{\text{10}}{\text{56}}=\frac{5}{\text{28}}$ .

4. $P\left(\text{At least one blue}\right)$

   By "at least one blue marble," we mean the following: one blue marble and two non-blue marbles, or two blue marbles and one non-blue marble, or all three blue marbles. So we have to find the sum of the probabilities of all three cases.

   $P\left(\text{At least one blue}\right)=P\left(\text{one blue, two non-blue}\right)+P\left(\text{two blue, one non-blue}\right)+P\left(\text{three blue}\right)$
   $P\left(\text{At least one blue}\right)=\frac{\mathrm{3C1}\times \mathrm{5C2}}{\mathrm{8C3}}+\frac{\mathrm{3C2}\times \mathrm{5C1}}{\mathrm{8C3}}+\frac{\mathrm{3C3}}{\mathrm{8C3}}$

   $P\left(\text{At least one blue}\right)=\text{30}/\text{56}+\text{15}/\text{56}+1/\text{56}=\text{46}/\text{56}=\text{23}/\text{28}$ .

   Alternately,

   we use the fact that $P\left(E\right)=1-P\left(E^c\right)$ .

   If the event $E=\text{At least one blue}$ , then $E^c=\text{None blue}$ .

   But from part c of this example, we have $\left(E^c\right)=5/\text{28}$

   Therefore, $P\left(E\right)=1-5/\text{28}=\text{23}/\text{28}$ .

Let us first do an easier problem–the probability of obtaining a pair of kings and queens.

Since there are four kings, and four queens in the deck, the probability of obtaining two kings, two queens and one other card is

To find the probability of obtaining two pairs, we have to consider all possible pairs.

Since there are altogether 13 values, that is, aces, deuces, and so on, there are $\text{13}\mathrm{C2}$ different combinations of pairs.