| 0 | 0.01 | 0.3 | 0.6 | 0.99 | 1 |
|---|
- This event is impossible. It can never occur.
- This event is certain; it will occur on every trial of the random phenomenon.
- This event is very unlikely, but it will occur once in a while in a long sequence of trials.
- This event will occur more often than not.
This implies that the probability is 0.
This implies that the probability is 1.
This implies that the probability is near zero but greater than zero; 0.01 satisfies this statement.
This implies that the probability is greater than one-half but less than one; 0.6 and 0.99 satisfy this statement.
- A gambler knows that red and black are equally likely to occur on each spin of a roulette wheel. He observes five consecutive reds occur and bets heavily on black at the next spin. Asked why, he explains that black is "due by the law of averages." Explain to the gambler what is wrong with this reasoning.
- After hearing you explain why red and black are still equally likely after five reds on the roulette wheel, the gambler moves to a poker game. He is dealt five straight red cards. He remembers what you said, and assumes that the next card dealt in the same hand is equally likely to be red or black. Is the gambler right or wrong, and why?
The wheel is not affected by its past outcomes; it has no memory. Outcomes are independent. So on any one spin, black and red remain equally likely.
Removing a card changes the composition of the remaining deck, so successive draws are not independent. If you hold 5 red cards, the deck now contains 5 fewer red cards (it has 26 black cards and 21 red cards), so your chance of another red card decreases.
The sum of the probabilities must equal one; therefore, the probability that Chavez is the one promoted can be found in the following manner:
P(Brown Promoted) + P(Chavez Promoted) + P(Williams Promoted) = 1
P(Chavez Promoted) = 1 - P(Brown Promoted) - P(Williams Promoted)
P(Chavez Promoted) = 1 - 0.25 - 0.2 = 1 - 0.45 = 0.55
| Probability | ||||
|---|---|---|---|---|
| Outcome | Model 1 | Model 2 | Model 3 | Model 4 |
 | 1/7 | 1/3 | 1/3 | 1 |
 | 1/7 | 1/6 | 1/6 | 1 |
 | 1/7 | 1/6 | 1/6 | 2 |
 | 1/7 | 0 | 1/6 | 1 |
 | 1/7 | 1/6 | 1/6 | 1 |
 | 1/7 | 1/6 | 1/6 | 1 |
In order for a probability model to be legitimate, the probabilities must be between 0 and 1 (inclusive) and the sum of the probabilities must be 1.
- Model 1 has each probability between 0 and 1; however, the sum of the probabilities is not 1 (it is actually 6/7). Therefore, Model 1 is not legitimate.
- Model 2 has each probability between 0 and 1, and the sum of the probabilities equals 1. Therefore, Model 2 is legitimate. (This model indicates that there is no "4" face and there are two "1" faces.)
- Model 3 has each probability between 0 and 1; however, the sum of the probabilities is not 1 (it is actually 7/6). Therefore, Model 3 is not legitimate.
- Model 4 has probabilities greater than 1, and the sum of the probabilities is not 1 (it is actually 8). Therefore, Model 4 is not legitimate.
The probability that an event does not occur is one minus the probability that the event does occur.
On any trial, the event either does or does not occur. So the percent of trials on which the event does not occur plus the percent of trials on which the event occurs is 100% (or 1.00), or all of the trials. Since we estimate probabilities with these relative frequencies, it is clear that the two probabilities must sum to 1 (or 100%). This gives us the following results:
P(Event) + P(No Event) = 1
P(No Event) = 1 - P(Event)
| 1 | $5000 prize | |
|---|---|---|
| 18 | $200 prizes | |
| 120 | $25 prizes | |
| 270 | $20 prizes |
What is the expected value of the winnings of one ticket in this lottery?
There are 1 + 18 + 120 + 270 = 409 winning tickets. Since there are 100,000 tickets, we see that there are 100,000 - 409 = 99,591 losing (or $0 prize) tickets. The expected value of a ticket is the sum of the prizes times their probabilities as follows:
Therefore, the expected value of a ticket is about 17 cents.
- What is the expected number of female offspring produced by a female Asian stochastic beetle? (See Example 12 in Section 7.2 for this insect's reproductive pattern [textbook page 355].)
The female Asian stochastic beetles have the following pattern of reproduction:
-
20% of females die without female offspring
30% have one female offspring
50% have two female offspring
The expected number of female offspring produced by a female Asian stochastic beetle is the sum of the number of offspring times their probabilities as follows:
Therefore, the expected number of female offspring is about 1.3 beetles.
| Age at death | 21 | 22 | 23 | 24 | 25 | 26 and up |
|---|---|---|---|---|---|---|
| Payout | -$19,900 | -$19,800 | -$19,700 | -$19,600 | -$19,500 | $500 |
| Probability | 0.0018 | 0.0019 | 0.0019 | 0.0019 | 0.0019 | 0.9906 |
The expected value of the earnings is the sum of the payouts times their probabilities as follows:
-
E(Earnings) = (-$19,900)(0.0018) + (-$19,800)(0.0019) + (-$19,700)(0.0019) + (-$19,600)(0.0019) +
(-$19,500)(0.0019) + ($500)(0.9906)
E(Earnings) = -$35.82 - $37.62 - $37.43 - $37.24 - $37.05 + $495.30 = $310.14
Therefore, the expected value of the earnings is about $310.14.
We will use the following coding for our simulation:
| Random Number | Simulated Result |
|---|---|
| 0, 1 | No Female Offspring (20%) |
| 2, 3, 4 | One Female Offspring (30%) |
| 5, 6, 7, 8, 9 | Two Female Offspring (50%) |
Using Table A (textbook page 430) and starting with line 101, we get the following values (random numbers above, simulated result below):
-
19223 95034 05756 28713 96409 12531 42544 82853
02111 22011 02222 12201 22102 01210 11211 21221
73676 47150 99400 01927 27754 42648 82425 36290
21222 12020 22100 00212 12221 11212 21112 12120
45467 71709 77558 00095
12122 20202 22222 00022
Therefore, we simulated a total of 130 female offspring from 100 female beetles. Our simulated expected number of offspring is 130/100 = 1.30. This is the same as the actual value of 1.3 that we calculated in Exercise 7.36(a). This supports the Law of Large Numbers by showing that with a large sample size, the estimated expected value approaches the "true" expected value.
Answers will vary according to each student's personal beliefs.
- The birth date of a randomly chosen person is equally likely to be any of the 365 dates of the year.
- The birth dates of different people in the room are independent.
Explain carefully how you would simulate the birth dates of 23 people to see if any two have the same birthday. Do the simulation once, using line 139 of Table A. (Comment: This simulation is most easily done by letting three-digit groups stand for the birthdates of successive people, so that 001 is January 1 and 365 is December 31. Ignore leap years. Some groups must be skipped in doing this. The simulation is too lengthy to ask you to repeat it many times, but in principle you can find the probability of matching birthdays by routine repetition. The birthday problem is too hard for most of your math-major friends to solve, so it shows the power of simulation.)
The birthday of a single person is simulated by a three-digit group of random digits. Groups 001 to 365 stand for the days of the year (ignoring those born on February 29), and others are ignored. By part (b) above, 23 consecutive three-digit groups (skipping those not between 001 and 365, inclusive) simulate the birth dates of 23 persons. Beginning in line 139, these are:
-
555|88 9|940|4 70|708 | 410|98 4|356|3 56|934 | 483|94 5|171|9
12|975 | 132|58 1|304|8 45|144 | 723|21 8|194|0 00|360 | 024|28
9|676|7 35|964 | 238|22 9|601|2 94|591 | 651|94 5|084|2 53|372 |
728|29 5|023|2 97|892 | 634|08 7|791|9 44|575 | 248|70 0|417|8
88|565 | 426|28 1|779|7 49|376 | 617|62 1|695|3 88|604 | 127|24
Note that the first two simulated people have the same birthdates! (No, I didn't cheat by trying several lines. The birthday problem is discussed in many tests on probability. With 23 persons, the probability of at least one match is 0.507.) In order to estimate the probability of at least two persons having the same birthday (in the same room of 23 persons), you would need to repeat the above simulation many times (perhaps 100 total times) and use the proportion of those samples that contained duplicate birthdays as your estimate of the probability of having at least two persons with the same birthday.
