7.6 Using Probability

The following examples will explore more applications of probability. Let us start by looking at flipping a coin, which is a single event. The total number of possible outcomes when flipping a coin is 2, either heads or tails. The two would be placed in the denominator. If we wanted to know the probability that the outcome would be heads then the expression would look like images or a 0.5% chance it would be heads.

Frequently, we want to know the probability of two things happening; in other words, one thing happens AND then another thing happens (AND means multiply). You multiply the probability of one thing happening by the probability of the other thing happening. What is the probability of the following:

1. Flipping two heads in a row? [Answer: images].
2. Flipping three heads in a row? [Answer: images].
3. Flipping six heads in a row? [Answer images] (rolling one die twice is the same as rolling two dice together once).
4. Drawing 2 aces? It depends on if you put the first one back before drawing the second. If you did put it back then it is 4/52 × 4/52 = 16/2704 or 1/169. If you did not put it back then it is 4/52 × 3/51 = 12/2652 or 1/221. The second probability is 3/51 because there are only 3 aces left and 51 cards left to choose from after you successfully take out the first ace. You have a better chance of getting 2, aces if you put the first one back before drawing again.
5. Throwing an 11 using 2 dice? [Answer: 2/36 because there are two ways of throwing an 11 {5 + 6 and 6 + 5}].

Now, look at one that is a little more complicated, such as rolling two dice. There are 36 possible outcomes when rolling two dice. Using Table 7.4, the probability that a three would be rolled could be determined.

Table 7.4 Dice Matrix

NumberTable

There are two possible ways of rolling the two dice so that they add up to three; therefore, the probability of rolling two dice so that they add up to three is 2/36 or 1/18 chances.

Using the matrix above, one could determine the possibility of rolling two dice and rolling a 3 thrice in a row. The probability of rolling a three was 1/9 as determined above. To find the probability of rolling a 3 thrice in a row, multiply 1/9 by itself thrice or 1/9 × 1/9 × 1/9 = 1/729.

Suppose you have a bag containing 14 red marbles, 12 blue marbles, and 18 green marbles.

1. What is the probability that if you pull out a marble at random you get either a red or a blue marble? [P(red or blue)] There are 44 total marbles and 26 chances to pull either a red or a blue marble, so the probability is 26/44 or 13/22.
2. What is the probability of not getting a blue marble? [P(not blue)] There are 44 total marbles and 32 are not blue, so the probability is 32/44 or 8/11.

Suppose you roll two dice, a red one and a green one.

1. What is P(a sum of 4)? There are three ways of making a sum of 4 (see matrix above) out of 36 possible sums, so the probability is 3/36 or 1/12.
2. What is P(a sum of 5 or 6 or 7)? There are four ways of making a sum of 5, five ways of making a sum of 6, and six ways of making a sum of 7, so the probability of making a sum of 5 or 6 or 7 is 4/36 + 5/36 + 6/36 = 15/36 or 5/12.

When a polling agency says its confidence level is 95%, it is saying that the probability of its numbers being correct is 0.95. What is the probability their numbers are wrong? (100% − 95%) or (1 − 0.95) = 5% or 0.05.

An art class has 13 right-handers and 7 left-handers.

1. What is the probability that a student chosen at random from the class is right-handed? There are 20 total students, so the probability that the student would be a right-hander is 13/20.
2. If three students are chosen, and the first two are right-handers, what is the probability that the third is also a right-hander? Two have already been picked and that leaves 18 students to choose the third student from and only 11 are right-handed since two have been chosen, so the probability that the student would be a right-hander is 11/18.

Tree diagrams can help you to visualize a probability problem such as flipping a coin or rolling dice. The diagram helps you to visualize a probability problem. Suppose you are going to flip a coin three times. You could see a tree diagram to visualize all of the possibilities. An example of this is in Figure 7.2.

Figure 7.2 Tree diagrams of coin flip.

7.2

So, one of the possibilities would be T, H, H. There are eight possible combinations when flipping a coin three times in a row.

1. What is the probability of getting three heads in a row (H, H, H)? There is only one branch that produces three heads in a row (H, H, H), so the probability is 1/8.
2. What is the probability of getting two heads and one tail (in any order)? There are three branches that have H, H, T (in any order), so the probability is 3/8.

What is the probability of being dealt two hearts? Solve this problem by using Figure 7.3, a tree diagram.

Figure 7.3 Tree diagram of being dealt two hearts.

7.3

To solve, multiply the probabilities along one whole “limb.” So multiplying along the top limb, we have 13/52 × 12/51 = 0.06 or 6%. With a tree diagram we can see other possible outcomes. For example, looking along the very bottom “limb” the probability of not getting a heart at all is 39/52 × 38/51 = 0.56 or 56%.

To summarize tree diagrams for probability

1. Conditional probabilities start at their condition.
2. Nonconditional probabilities start at the beginning of the tree.
3. Multiply when moving horizontally across a limb.
4. Add when moving vertically from limb to limb.
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