Because many of the examples in this chapter use dice, coins and playing cards, their characteristics are reviewed here:
Dice
The standard die (the singular of dice) used in the Western world is a cube with six sides, each displaying a different number of dots, from 1 to 6.A stan- dard assumption in probability calculations is that all sides of the die are equally likely to land facing up when the die is rolled or thrown, so one roll of the die has six equally likely outcomes: 1, 2, 3, 4, 5, and 6.In technical terms, the set of outcomes from rolling one or more dice has a discrete uniform distribution because the possible outcomes can be enumerated and each outcome is equally likely.The results of two or more dice thrown at once (or multiple throws of the same die) are assumed to be independent of each other, so the probabilities of each combination of numbers are calculated by multiplying the probability of each separate result.
In the interests of precision, I should point out that the “equal probability for all sides” holds only for casino dice, in which the pips (circles used to mark the numbers on each side) are painted on.Cheaper dice, such as you may purchase at the dime store, do not have equal weight on all sides because the pips are drilled into the cube face rather than painted on.However, in theo- retical discussions of probability, this nicety is usually ignored and we assume that all sides of the dice are equally probable.
Coins
The archetypal coin used in probability experiments has two sides, heads and tails.A fair coin is equally likely to come up heads or tails on any toss or flip. For any coin, fair or not, the probability of heads and tails is constant on each flip, so that the results of previous flips have no influence on later flips and the results of multiple flips are independent of each other.As with dice, the probability of an actual coin landing heads or tails is seldom exactly 50–50, for a number of physical reasons, including coin design and wear, and off- center technique on the part of the person performing the flip, but for the sake of probability exercises we assume it is unless otherwise specified.Some- times experiments are conducted by spinning coins rather than flipping them (fewer projectiles flying through the air in a crowded classroom).However, the 50–50 assumption applies even less here, although for the purposes of doing calculations (as opposed to actually spinning coins and recording the results) we assume that it does.For more on these issues, see http://www.
sciencenews.org/articles/20040228/fob2.asp.
Playing cards
The standard deck of playing cards today has 52 cards in four suits: spades, clubs, diamonds, and hearts.Spades and clubs are black cards, diamonds and hearts are red cards.There are 13 cards in each suit: an ace, numbered cards from 2 through 10, and 3 face cards—the jack, queen, and king.
Exercises
Problem
If I draw one card from an ordinary deck of 52 playing cards, what is the proba- bility that it will be a red card?
Solution
1. The trial is a single draw of one card from a deck of 52.
2. The sample space is all the possible cards, each of which has an equal like- lihood of being drawn.
3. The event isE = {red card}.
4. Since there are 52 cards in the deck and half (26) are red, the probability of drawing a red card is 26/52 or 0.5. The answer is that we have a 50% proba- bility of drawing a red card.
Problem
If I roll a die once, what is the probability of getting a number lower than 5? Solution
1. The trial is a single roll of a six-sided die.
2. The sample space is the numbers (1, 2, 3, 4, 5, 6), all of which are equally likely.
3. The event isE= (any of 1, 2, 3, 4), which can also be considered the union of four simple events, i.e.,E = (E = 1)∪ (E = 2)∪ (E = 3)∪ (E = 4).
4. Four of the six simple events or possible outcomes that constitute the sample space satisfy the eventE, so the probability ofE is 4/6 or 0.67 (rounded). Alternative solution
Another way to look at this is to calculate the probability of each simple event that satisfies the eventEand add them together, since the events are mutually exclu- sive.Using this approach, the probability of each simple event inEis 1/6, i.e., there is a 1 in 6 chance that the number will be 1, 1 in 6 that the number will be 2, and so on, so the probability ofEis 1/6 + 1/6 + 1/6 + 1/6 or 4/6, which is the same answer as above.
Problem
If I flip a fair coin twice, what is the probability that I will get at least one head? Solution
1. The experiment is two flips of a fair (P= 0.5 for either heads or tails) coin, i.e., two independent trials.
Exercises | 37
Probability
3. The event isE= (at least one head); three of the events in the sample space satisfy this condition: (h, h), (h, t), and (t, h).
4. Each of the outcomes is equally likely, and three of the four satisfy the event E, so the probability ofE is 3/4 or 0.75.
Alternative solution
We can also find this result mathematically by calculating the probability of the complement of this event, then subtracting it from 1 to get the probability of the event.If the eventEis (at least one head) then its complement is ~E= (no heads, i.e., two tails). We know that the probability of getting a tail on any flip of a fair coin is 0.5, and the flips are independent, so the probability of (t,t) is 0. 5×0.5 or 0.25. Using the definition of complement above, 1 –P(~E) =P(E), so 1 – 0.25 = 0.75 orP(E) and the probability of at least one head from two flips is 0.75. Problem
If I draw one card from a standard 52-card deck, what is the probability that it will be a black (clubs or spades) face card (king, queen, or jack)?
Solution
1. The trial is drawing one card from a 52-card deck.
2. The sample space is all 52 cards, each of which has equal probability of being drawn.
3. The event isE= (black face card); six cards satisfy this condition, the jack, queen, or king of either spades or clubs.
4. The probability is 6/52 or 0.115. Mathematical solution
P(face card) = 12/52 or 0.231 P(black card) = 26/52 or 0.5
P(black face card) =P(face card) *P(black card) = 0.231 * 0.5 = 0.116
This solution is possible because the probability of drawing a black card, and the probability of drawing a face card are independent.
Problem
If I draw one card from a standard 52-card deck, what is the probability that it will be either black (clubs or spades) or a face card (king, queen, or jack)?
Solution
1. The trial is drawing one card from a 52-card deck.
2. The sample space is all 52 cards, each of which has an equal probability of being drawn.
3. The event isE= (either black card or face card), meaning any of the 26 black cards or any of the 12 face cards will satisfy the event.
4. The two types of cards that will satisfy the condition are not mutually exclu- sive: some black cards are also face cards, and vice versa.There are 26 black cards: ace through king of spades (13) and ace through king of clubs (13). There are 12 face cards: jack, king, and queen for each of hearts, diamonds, clubs, and spades.There are six cards in both categories: jack, king, and queen of spades and clubs.So there are 26 + 12 – 6 = 32 cards that satisfy this event, and the probability is 32/52 or 0.615.
Mathematical solution
P(black card) = 26/52 or 0.500 P(face card) = 12/52 or 0.2301 P(black face card) = 6/52 or 0.115
P(black card or face card) = 0.500 + 0.231 – 0.115 = 0.616 Problem:
If I draw a single card from a 52-card deck and it is black, what is the probability that its suit is clubs?
Solution
1. The trial is drawing one card from a 52-card deck.
2. The sample space is all black cards, since we are interested in the conditional probability of a card being a club, given that it is a black card.Our sample space is therefore the 26 black cards.
3. The event isE = (club | black card).
4. The probability of the card being a club, given that it is a black card, is 13/26 or 0.5.
Note that this is aconditionalprobability (conditioned on the fact that the card is black); theunconditionalprobability of the card being a club, if we had no information about its color, is 13/52 or 0.25.
Mathematical solution
P(clubs | black card) =P(clubs)/P(black card) = 0.25/0.5 = 0.5 Problem
If order is not significant, how many ways are there to select a subset of 5 students from a classroom of 20?
Solution
This is a combinatorial problem that is too lengthy to solve by listing all possible subsets. Instead, we will use the combination formula. In this case,n = 20 andk = 5:
Exercises | 39
Probability
Problem
There are 80 students attending a conference: 40 boys and 40 girls.30 of the boys are majoring in math, as are 20 of the girls.We know that if you pick a boy at random, there is a 75% chance that he is a math major; however, if you pick a math major at random, what is the probability that the student is male?
Solution
P(male) = 40/80 = 0.5 P(~male) = 40/80 = 0.5 P(math|male) = 30/40 = 0.75 P(math|~male) = 20/40 = 0.5