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     Probability questions are becoming increasingly common. Probability questions tend to be bundled among the difficult questions, so high scorers will commonly encounter them. If you are a low scorer and are pressed for time, consider skipping most of the material past "Simple Probability." This is a computer-adaptive test, and low scorers aren't likely to encounter the most difficult probability question types.


A. Simple Probability
B. Probability of Multiple Events
C. Independent and Dependent Events
D. Mutually Exclusive Events
E. Conditional Probabilities
F. Combinations

      

A. Simple Probability

     In general, the probability of an event is the number of favorable outcomes divided by the total number of possible outcomes.

      Probability= (# of favorable outcomes) / (# of possible outcomes)

 

Example 1

What is the probability that a card drawn at random from a deck of cards will be an ace?

Solution
In this case there are four favorable outcomes:
(1) the ace of spades
(2) the ace of hearts
(3) the ace of diamonds
(4) the ace of clubs.

Since each of the 52 cards in the deck represents a possible outcome,
there are 52 possible outcomes. Therefore, the probability is 4/52 or 1/13.

The same principle can be applied to the problem of determining the probability of obtaining different totals from a pair of dice.

Example 2

What is the probability that when a pair of six-sided dice are thrown, the sum of the numbers equals 5?

Solution
There are 36 possible outcomes when a pair of dice is thrown. Consider that if one of the dice rolled is a 1, there are six possibilities for the other die. If one of the dice rolled a 2, the same is still true. And the same is true if one of the dice is a 3,4,5, or 6. If this is still confusing, look at the following (abbreviated) list of outcomes: [(1,1),(1,2),(1,3),(1,4),(1,5),(1,6); (2,1),(2,2),(2,3)… (3,1),(3,2),3,3)… (4,1)…(5,1)…(6,1)….

The total number of outcomes is 6 × 6 = 36. Since four of the outcomes have a total of 5 [(1,4),(4,1),(2,3),(3,2)], the probability of the two dice adding up to 5 is 4/36 = 1/9.

Example 3

What is the probability that when a pair of six-sided dice are thrown, the sum of the number equals 12?

Solution
We already know the total number of possible outcomes is 36, and since there is only one outcome that sums to 12, (6,6--you need to roll double sixes), the probability is simply 1/36.

 

NOTE: The material from here on through the end of the section is dense and intended only for medium to high scorers. Because this is a CAT (computer-adaptive test), it is relatively unlikely that lower scorers will encounter these questions, and, if they are short of time, they are better off putting their time into other sections.

B. Probability of Multiple Events

    For questions involving single events, the formula for simple probability is sufficient. For questions involving multiple events, the answer combines the probabilities for each event in ways that may seem counter-intuitive. The following strategy is excellent for acquiring a better feel for probability questions involving multiple events or for making a quick guess if time is short. We will focus on questions involving two events.

• If two events have to occur together, generally an "and" is used. Take a look at Statement 1: "I will only be happy today if I get email and win the lottery." The "and" means that both events are expected to happen together.
  
  
• If both events do not necessarily have to occur together, an "or" may be used as in Statement 2, "I will be happy today if I win the lottery or have email."
  
  

    Consider Statement 1. Your chances of getting email may be relatively high compared to your chances of winning the lottery, but if you expect both to happen, your chances of being happy are slim. Like placing all your bets at a race on one horse, you've decreased your options, and therefore you've decreased your chances. The odds are better if you have more options, say if you choose horse 1 or horse 2 or horse 3 to win. In Statement 2, we have more options; in order to be happy we can either win the lottery or get email.

     The issue here is that if a question states that event A and event B must occur, you should expect that the probability is smaller than the individual probabilities of either A or B. If the question states that event A or event B must occur, you should expect that the probability is greater than the individual probabilities of either A or B. This is an excellent strategy for eliminating certain answer choices.

These two types of probability are formulated as follows:

Probability of A and B
P(A and B) = P(A) × P(B).

In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B.

Probability of A or B
P(A or B) = P(A) + P(B).

In other words, the probability of A or B occurring is the sum of the probability of A and the probability of B (this assumes A + B cannot both occur). If there is a probabiilty of A and/or B occuring, then you must subtract the overlap.

Look at the following examples.

Example 4

If a coin is tossed twice, what is the probability that on the first toss the coin lands heads and on the second toss the coin lands tails?
a) 1/6
b) 1/3
c) ¼
d) ½
e) 1

Solution
First note the "and" in between event A (heads) and event B (tails). That means we expect both events to occur together, and that means fewer options, a less likely occurrence, and a lower probability. Expect the answer to be less than the individual probabilities of either event A or event B, so less than ½. Therefore, eliminate d and e. Next we follow the rule P(A and B) = P(A) × P(B). If event A and event B have to happen together, we multiply individual probabilities. ½ × ½ = ¼. Answer c is correct.

NOTE: Multiplying probabilities that are less than 1 (or fractions) always gives an answer that is smaller than the probabilities themselves.

 

Example 5

If a coin is tossed twice what is the probability that it will land either heads both times or tails both times?
a)1/8
b)1/6
c)1/4
d)1/2
e)1

Solution
Note the "or" in between event A (heads both times) and event B (tails both times). That means more options, more choices, and a higher probability than either event A or event B individually. To figure out the probability for event A or B, consider all the possible outcomes of tossing a coin twice: heads, heads; tails, tails; heads, tails; tails, heads. Since only one coin is being tossed, the order of heads and tails matters. Heads, tails and tails, heads are sequentially different and therefore distinguishable and countable events. We can see that the probability for event A is ¼ and that the probability for event B is ¼. We expect a greater probability given more options, and therefore we can eliminate choices a, b and c, since these are all less than or equal to ¼. Now we use the rule to get the exact answer. P(A or B) = P(A) + P(B). If either event 1 or event 2 can occur, the individual probabilities are added: ¼ + ¼ = 2/4 = ½. Answer d is correct.

NOTE: We could have used simple probability to answer this question. The total number of outcomes is 4: heads, heads; tails, tails; heads, tails; tails; heads, while the desired outcomes are 2. The probability is therefore 2/4 = ½.

The following chart summarizes the "and's" and "or's" of probability:

Probability	Formula	Expectation
P(A and B)	P(A) × P(B)	Lower than P(A) or P(B)
P(A or B)	P(A) + P(B)	Higher than P(A) or P(B)

C. Independent and Dependent Events

     The types of events that we have discussed so far are all independent events. By independent we mean that the first event does not affect the probability of the second event. Coin tosses are independent. They cannot affect each other's probabilities; the probability of each toss is independent of a previous toss and will always be 1/2. Separate drawings from a deck of cards are independent events if you put the cards back. An example of a dependent event, one in which the probability of the second event is affected by the first, is drawing a card from a deck but not returning it. By not returning the card, you've decreased the number of cards in the deck by 1, and you've decreased the number of whatever kind of card you drew. If you draw an ace of spades, there are 1 fewer aces and 1 fewer spades. This affects our simple probability: (number of favorable outcomes)/ (total number of outcomes. This type of probability is formulated as follows:

If A and B are not independent, then the probability of A and B is

P(A and B) = P(A) × P(B|A)

where P(B|A) is the conditional probability of B given A.

 

Example 6

If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces?

Solution
Event A is that the first card is an ace. Since 4 of the 52 cards are aces, P(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17, and the probability of A and B is 1/13 × 1/17 = 1/221. The same reasoning is applied to marbles in a jar.

Example 7

If there are 30 red and blue marbles in a jar, and the ratio of red to blue marbles is 2:3, what is the probability that, drawing twice, you will select two red marbles if you return the marbles after each draw?

Solution
First, let's determine the number of red and blue marbles respectively. The ratio 2:3 tells us that the total of 30 marbles must be broken into 5 groups of 6 marbles, each with 2 groups of red marbles and 3 groups of blue marbles. Setting up the equation 2x + 3x = 5x =30 employs the same reasoning. Solving, we find that there are 12 red marbles and 18 blue marbles. We are asked to draw twice and return the marble after each draw. Therefore, the first draw does not affect the probability of the second draw. We return the marble after the draw, and therefore, we return the situation to the initial conditions before the second draw. Nothing is altered in between draws, and therefore, the events are independent.

Now let's examine the probabilities. Drawing a red marble would be 12/30 = 2/5. The same is true for the second draw. Since we want two red marbles in a row, the question is really saying that we want a red marble on the first draw and a red marble on the second draw. The "and" means we should expect a lower probability than 2/5. Understanding that the "and" is implicit can help you eliminate choices d and e which are both too big. Therefore, our total probability is P(A and B) = P(A) ×. P(B) = 2/5 × 2/5 = 4/25.


Now consider the same question with the condition that you do not return the marbles after each draw. The probability of drawing a red marble on the first draw remains the same, 12/30 = 2/5. The second draw, however, is different. The initial conditions have been altered by the first draw. We now have only 29 marbles in the jar and only 11 red. Don't panic! We simply use those numbers to figure our new probability of drawing a red marble the second time, 11/29. The events are dependent and the total probability is P(A and B) = P(A) ×. P(B) = 2/5 × 11/29 = 132/870 = 22/145.

If you return every marble you select, the probability of drawing another marble is unaffected; the events are independent. If you do not return the marbles, the number of marbles is affected and therefore dependent.

 

D. Mutually Exclusive Events

     Another type of probability deals with mutually exclusive events. What do we mean by mutually exclusive events? And what does it mean for two events not to be mutually exclusive? Consider the following example of drawing cards:

Example 8

What is the probability that a card selected from a deck will be either an ace or a spade?
a)2/52
b)2/13
c)7/26
d)4/13
e)17/52

Solution
First, identify events A and B and notice the "or" in between them. That means a greater probability than either A or B individually. Therefore, we expect the answer to be greater than 4/52(ace)=1/13 or 13/52(spade)= 1/4. Eliminate a and b. The tricky part of this question lies in the fact that when we figure probability, we are really just counting, and sometimes, we count twice. In this case we have counted the ace of spades twice. If you don't see this, consider what the 4 in 4/52 stands for: ace of hearts, ace of diamonds, ace of clubs, ace of spades. The 13 in 13/52 stands for all the spades: 1,2,3…King, Ace(of spades). Therefore if we just combined the probabilities by the rule for P(A or B) = P(A) + P(B) we would be over counting. We have to subtract 1/52, the ace of spades that was counted twice. Our answer becomes 4/52 + 13/52 - 1/52 = 16/52 = 4/13.

Another way to think about the question is to just count aces and spades; that is, use simple probability. There are 13 spades in a deck and 3 aces other than the ace of spades already included in the 13 spades. Therefore, there are 16 desired outcomes out of a total of 52 possible outcomes, or 16/52 = 4/13.

In the above example, events A and B are not mutually exclusive. Figuring the probability for event A includes part of the probability of event B, and we must therefore subtract out this "over-counted" probability to get the correct answer.

The following example illustrates mutually exclusive events:

Example 9

What is the probability that a card selected from a deck will be either an ace or a king?
a)1/169
b)1/26
c)2/13
d)4/13
e)8/13

Solution
The question asks for either an ace or a king. Since there are four kings and four aces in a deck, the probabilities for event A and event B are the same, 4/52 = 1/13. Our answer must be more than this, so eliminate a and b. Do kings and aces have anything to do with each other? Is there such a thing as an ace of kings or a king of aces? No, so we don't have to worry about having over-counted; the events are mutually exclusive. The probability is straightforward: P(A or B) = P(A) + P(B) = 1/13 + 1/13 = 2/13. C is correct.

Again we could have used simple probability. Count the total number of kings and aces (4+4) and divide by the total number of cards in a deck: 8/52 = 2/13.

 

E. Conditional Probabilities


A conditional probability is the probability of an event given that another event has occurred.

Example 10

What is the probability that the total of two dice will be greater than 8 given that the first die is a 6?

Solution
This can be computed by considering only outcomes for which the first die is a 6. Then, determine the proportion of these outcomes that total more than 8. All the possible outcomes for two dice are shown in the section on simple probability. There are 6 outcomes for which the first die is a 6: (6,1),(6,2),(6,3),(6,4),(6,5),(6,6), and of these, there are four that total more than 8. The probability of a total greater than 8 given that the first die is 6 is therefore 4/6 = 2/3.

 

1. Probability of A and B

    If A and B are independent, then the probability that events A and B both occur is p(A and B) = p(A) × p(B). In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B. What is the probability that a coin will come up with heads twice in a row? Two events must occur: a heads on the first toss and a heads on the second toss. Since the probability of each event is 1/2, the probability of both events is: 1/2 × 1/2 = 1/4. Now consider a similar problem: someone draws a card at random out of a deck, replaces it, and then draws another card at random. What is the probability that the first card is the ace of clubs and the second card is a club (any club)?
     Since there is only one ace of clubs in the deck, the probability of the first event is 1/52. Since 13/52 = 1/4 of the deck is composed of clubs, the probability of the second event is 1/4. Therefore, the probability of both events is 1/52 × 1/4 = 1/208.
       What's the probability of A and B (2 of 2) if A and B are not independent? If A and B are not independent, then the probability of A and B is p(A and B) = p(A) × p(B|A) where p(B|A) is the conditional probability of B given A. If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces? Event A is that the first card is an ace. Since 4 of the 52 cards are aces, p(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17, and the probability of A and B is 1/13 × 1/17 = 1/221.

 

2. Probability of A or B

     If events A and B are mutually exclusive, then the probability of A or B is simply:
p(A or B) = p(A) + p(B).

     What is the probability of rolling a die and getting either a 1 or a 6? Since it is impossible
to get both a 1 and a 6, these two events are mutually exclusive. Therefore,

p(1 or 6) = p(1) + p(6) = 1/6 + 1/6 = 1/3

If the events A and B are not mutually exclusive, then

p(A or B) = p(A) + p(B) - p(A and B).

The logic behind this formula is that when p(A) and p(B) are added, the occasions on
which A and B both occur are counted twice. To adjust for this, p(A and B) is subtracted.



Example 11

What is the probability that a card selected from a deck will be either an ace or a spade?



Solution
The relevant probabilities are
p(ace) = 4/52
p(spade) = 13/52

The only way an ace and a spade can both be drawn is to draw the ace of spades. There is only one ace of spades, so p(ace and spade) = 1/52. The probability of an ace or a spade can be computed as p(ace)+p(spade)-p(ace and spade) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.

F. Combinations

       Suppose that a job has two different parts. There are m different ways of doing the first part, and there are n different ways of doing the second part. The problem is to find the number of ways of doing the entire job. For each way of doing the first part of the job, there are n ways of doing the second part. Since there are m ways of doing the first part, the total number of ways of doing the entire job is m x n. The formula that can be used is

Number of ways = m x n

      For any problem that involves two actions or two objects, each with a number of choices, and asks for the number of combinations, the above formula can be used.

Example 13

William wants a sandwich and a drink for lunch. If a restaurant has 4 choices of sandwiches and 3 choices of drinks, how many different ways can he order his lunch?

Solution
Since there are 4 choices of sandwiches and 3 choices of drinks, using the formula

Number of ways = 4(3) = 12

Therefore, the man can order his lunch 12 different ways.

Now, what if the combinations available decrease after each combination is taken?

Example 14

There are five meal options in the cafeteria of a certain school. Assuming that a different meal must be eaten each day, and each different type of meal must be eaten once before any type of meal can be eaten a second time, how many different orders of meals can a student eat in the first five days?


Solution
The answer is 120. There are five types of meals, so the total number of possibilities is 5!. (the "!" stands for factorial), or 5 x 4 x 3 x 2 x 1 = 120. Since a different meal is assigned to every day, you must reduce the amount that you multiply by on a daily basis from 5 to 4 to 3 to 2 to 1. If you like, assign the letters A, B, C, D and E to the five meals and see how many different orders you can create.

What if two or more samples are chosen at a time? If we have objects a, b, c, d and want to arrange them two at a time--that is, ab, bc, cd, etc. (We have four combinations taken two at a time). If you have four different combinations taken two at a time, you can write this as 4 C 2, which can be written as

4 C 2 = 4 x 3
            2 x 1

Examples

8 C 3 = 8 x 7 x 6
            3 x 2 x 1

10 C 4 = 10 x 9 x 8 x 7
              4 x 3 x 2 x 1

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Note#2: You cannot calculate (1/6) (5/6) + (5/6) (1/6) which undercounts the possiblities. Simply add combinations where it cannot be true.