Ch. 5_Probability

Probability of success + Probability of failure =

1

The greatest probability =

1 (the certainty that an event will occur). The probability of 1 means that the event must occur.

The probability of the events *not* occurring =

1 - x In certain types of *OR* questions, the probability of the event not occurring may be easier to calculate.

You should handle OR problems in one of two ways, depending on whether the events can occur together in the same scenario:

1. If the events *cannot* occur together. 2. If the events *can* occur together.

1. If the events *cannot* occur together.

If an *OR* problem features events that *cannot* occur together, then you can find the OR probability by *adding* the probabilities of the individual events.

2. If the events *can* occur together.

If an OR problem features events that *can* occur together, then use this formula to find the OR probability: P(A *or* B) = P(A) + P(B) - P(A *and* B) You need to subtract the probability of A and B (i.e., P(A *and* B)) or else you are overestimating the probability that A and B occurring.

And =

Multiply *To get a lower number (i.e., lower probability of success)

Probability =

Number of *desired or successful* outcomes/Total number of *possible* outcomes

Example: What is the probability that none of the rolls will yield a 6?

On each roll there is a 5/6 probability that the die will not yield a 6. Thus, the probability on all 3 rolls that the die will not yield a 6 = 5/6 x 5/6 x 5/6 = 125/216 = 1 - 125/216 = 91/216

The Domino Effect

Sometimes the outcome of the first event will affect the probability of a subsequent event.

Example: If a fair coin is tossed three times, what is the probability that it will turn up heads exactly twice?

Step 1: You have to formulate equally likely outcomes in terms of the outcome of each flip. Step 2: Formulate a counting tree, which breaks down possible outcomes step by step, with only one decision at each branch of the tree. HHH HHT HTH THH HTT THT TTH TTT The outcomes on this list (HHT, HTH, THH) have heads exactly twice so the probability of exactly two heads is 3/8. Can also be written: P(exactly 2 heads) = 3/8

The lowest probability =

0/6 = 0 Expressed as 0%.

OR =

Add *To get a larger number (i.e., larger probability of success)

15. A zoo has w wildebeests, y yaks, z zebras, and no other animals. If an animal is chosen at random from he zoo, is the probability of choosing a yak greater than the probability of choosing a zebra? 1. y/(w + z) > 1/2 2. z/(w + y) < 1/2

C. Guessed. Got right. Question: Rephrase: Are there more yaks than zebras? Or Is y > z? 1. Cross-multiply to get: 2y > w + z However, we do not know whether y is greater than z. For instance, if y = 5 and w = 1, z could be either 8 or 1 (or other values). Not sufficient. 2. Cross-multiply to get: 2z < w + y Thus, we do not know whether y is greater than z. If y = 3 and w = 7, z could be either 4 or 1 (or other values). C. Use cross-multiplication. Flip one of them to point the inequality in the same direction, then add them up: w + z < 2y 2z < w + y = w + 3z < w + 3y = 3z < 3y = z < y I said: Question: Is 1/y > 1/z? Was not sure we can cross-multiply. 1. No rephrase. 2. No rephrase. C. z/w + y < 1/2 < y/w + z

The Domino Effect Example: In a box with 10 blocks, 3 of which are red, what is the probability of picking out a red box on each of your first two tries? Assume that you do not replace the first block after you have picked it.

Probability of picking a red block on your first pick = 3/10 Probability of picking a red block on your second pick = 2/9 *Since you are taking away on block*. Thus, the probability of picking a red block on both picks = 3/10 x 2/9 = 6/90 = 1/15 *If you are suppose to replace the object, the problem should clearly tell you so.

Probability problems that deal with multiple events usually involve two operations:

1. Multiplication 2. Addition

The probabilities can be:

1. Percents between 0% to 100%, inclusive. 2. Fractions between 0 and 1, inclusive.

The probability of rolling a 5 on a dice =

1/6 because 5 corresponds to only *one* of those outcomes.

3. What is the probability that the sum of two dice will yield a 10 or lower?

11/12. I said 17/18. Careless. Calculate the probability that the sum will be higher than 10 and subtract the result from 1. There are 3 combinations of 2 dice that yield a sum higher than 10: 5 + 6, 6 + 5, 6 + 6. Therefore, the probability that the sum will be higher than 10 is 3/36, or 1/12. The probability that the sum will be 10 or lower is 1 - 1/12 = 11/12.

8. In a diving competition, each diver has a 20% chance of a perfect dive. The first perfect dive of the competition, but no subsequent dives, will receive a perfect score. If Janet is the third diver to dive, what is her chance of receiving a perfect score?

12.8%. I said .008. In order for Janet to receive a perfect score, neither of the previous two divers can receive one. Therefore, we are finding the probability of a chain of three events: that diver one will *not* get a perfect score AND diver two will *not* get a perfect score AND Janet *will* get a perfect score. Multiply the probabilities: 8/10 x 8/10 x 2/10 = 128/1000 = 12.8%.

10. 5 actresses are vying for 3 leading roles. The actresses are Julia, Meryl, Sally, Lauren, and Hallie. Assuming that no actress has any advantage in getting any role, what is the probability that Julia and Hallie will star in the film together?

13/18. I said 2/9, 1/3, and 4/9. Solve this problem by finding the probability that the two flowers in the boutique *will* be the same, and then subtract the result from 1. There are 10 different boutiques in which both flowers are the same. Then, find the number of different 2-flower boutiques that can be made in total, using an anagram model. In how many different ways can arrange the letters in the "word" YYNNNNNN? 9!/7!2! = 9 x 8/2 x 1 = 36 The probability of randomly putting together a bouquet that contains two of the same type of flower is 10/36 or 5/18. Therefore, the probability of randomly putting together a bouquet that contains two different flowers and that therefore will *not* need to be changed is 1 - 5/18, or 13/18.

1. What is the probability that the sum of two dice will yield a 4 or 6?

2/9. I said 1/3. There are 36 ways in which 2 dice can be thrown (6 x 6 = 36). The combinations that yield sums of 4 or 6 are: 1 + 3 2 + 2 and so on. There are 8 different combinations. Therefore, the probability is 8/36 or 2/9. I said: 1/6 + 1/6 = 2/6 = 1/3. I didn't treat "sum" correctly.

9. In a bag of marbles, 3 are red, 2 are white, and 5 are blue. If Bob takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marble?

2/9. I said 1/6. You can solve this by listing the winning scenarios or by using combinatorics counting methods. First pick: Blue = 1/2 White = 1/5 Second pick: White = 2/9 Blue = 5/9 Probability: Blue = 1/2 x 2/9 = 1/9 White = 1/5 x 5/9 = 1/9 OR 10!/(2!)(8!) = 45 Since there are 2 white marbles and 5 blue marbles, there are 2 x 5 = 10 different combinations. Therefore, 10/45 = 2/9

14. For one roll of a certain die, the probability of rolling a two is 1/6. If this die rolled 4 times, which of the following is the probability that the outcome will be a two at least 3 times? (1/6)^4 2(1/6)^3 + (1/6)^4 3(1/6)^3 (5/6) + (1/6)^4 4(1/6)^3 (5/6) + (1/6)^4 6(1/6)^3 (5/6) + (1/6)^4

4(1/6)^3 (5/6) + (1/6)^4. I guessed. I said 1/216. We must express the probability directly. We must regard the desired outcome in two separate parts: first, rolling a two exactly 4 times, and second, rolling a two exactly 3 times out of 4 attempts. First, the probability of rolling a two exactly 4 times is (1/6)(1/6)(1/6)(1/6) = (1/6)^4. Next, if you roll a two exactly 3 times out of 4 attempts, then on exactly of those attempts, you do *not* roll a two. Hence, the probability of rolling a two exactly 3 times out of 4 attempts is the sum of the following four probabilities: (two)(two)(two)(not two) → (1/6)(1/6)(1/6)(5/6) = (1/6)^3(5/6) (two)(two)(not two)(two) → (1/6)(1/6)(5/6)(1/6) = (1/6)^3(5/6) (two)(not two)(two)(two) → (1/6)(5/6)(1/6)(1/6) = (1/6)^3(5/6) (not two)(two)(two)(two) → (5/6)(1/6)(1/6)(1/6) = (1/6)^3(5/6)

The greatest probability example: # of successful/Total # of possible outcomes

6/6 = 1 The certainty is expressed as 100%.

6. There is a 30% chance of rain and 70% chance of shine. If it rains, there is a 50% chance that Bob will cancel his picnic, but if the sun is shining, he will definately have his picnic. What is the chance that Bob will have his picnic?

85%. I did not know. I used probability trees. There are two possible chains of events in which Bob will have his picnic: 1. Sun shine: P = 7/10 = 14/20 2. It rains AND Bob chooses to have the picnic: P = (3/10)(1/2) = 3/20. Add the probabilities together to find the total probability that Bob will have the picnic: 14/20 + 3/20 = 17/20 = 85%

13. In Town Z, the town lion roars on some days and not on others. If a day is chosen at random from last April, what is the probability that on that day, either the town lion roared or it rained? 1. Last April, the lion never roared on a rainy day. 2. Last April, the lion roared on 10 fewer days than it rained.

E. Guessed. I said A. Recognize that you are asked for an *OR* probability: P(Roar or Rain) = P(Roar) + P(Rain) - P(Roar and Rain). 1. P(Roar and Rain) = 0. Not sufficient. 2. Rainy days - 10 = Roaring days. Dividing by 30 (number of days in April). Rainy days/30 days = 10/30 = Roaring days/30 days P(Rain) - 1/3 = P(Roar) P(Rain) - P(Roar) = 1/3 C. We cannot fill in the equation for P(Roar or Rain). If statement (2) gave us P(Rain) + P(Roar), we could find P(Roar or Rain), but we are given the difference of the probabilities P(Rain) and P(Roar), not their sum. I said: Question: I made probability trees. 1. P(Roar, Rain) = 0 2. R = Rain - 10

Useful tool to keep track on branching possibilities and winning scenarios is

Probability Trees However, avoid setting up complicated trees. GMAT problems almost never require this. Use trees to conceptualize a path through the problem.

Example: A fair die is rolled once and a fair coin is flipped once. What is the probability that either the die will land on 3 or that the coin will land on heads?

These outcomes are not mutually exclusive, since both events can occur together. The probability that the die will land on 3 = 1/6 The probability that the coin will land on heads = 1/2. Therefore, the probability of either events occurring: = 1/6 + 1/2 - (1/6 x 1/2) = 7/12

1. Assume that X and Y are *independent* events. Example: What is the probability that a fair coin flipped twice will land on heads both times?

This is an *AND* problem. 1/2 x 1/2 = 1/4 When multiplied together, you get a smaller result, which means *lower* probability.

2. Assume that X and Y are *mutually exclusive events* (meaning that the two events cannot both occur). Example: What is the probability that a fair die rolled once will land on either 4 or 5?

This is an *OR* problem. The die cannot land on both 4 and 5 at the same time. 1/6 + 1/6 = 2/6 = 1/3

1. Assume that X and Y are *independent* events.

To determine the probability that event X *AND* event Y will both occur, *MULTIPLY* the two probabilities together.

2. Assume that X and Y are *mutually exclusive events* (meaning that the two events cannot both occur).

To determine the probability that event X *OR* event Y will occur, *ADD* the two probabilities together.

For the probability fraction to be meaningful

all the outcomes must be equally likely.

You will have to think carefully how to

break a situation down into equally likely outcomes.

If on a GMAT problem, "success" contains multiple possibilities - especially if the wording contains phrases such as "at least" and "at most" - then

consider finding the probability that success does not happen. If you can find this failure probability more easily (i.e., x) then the probability you really want to find is 1 - x.

Probability measures

how often an event will occur in a long series of repeated trials.

You may see OR problems

in disguise Such problems ask for the probability of what seems to be a single event.