Toss 3 coins. Find the probability that at least one head shows by writing out the sample space

Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}; probability of at least one head is then $7/8=0.875$Is it unusual to get 3 heads when 3 coins are tossed?

The probability of 3 heads is $1/8 = 0.125$, which is not terribly unusual. (*Typically, events that have probability less than $0.05$ are considered unusual.*)Answer the following:

- What is the probability of an event that must happen?
- What is the probability of an event that can't happen?
- A sample space consists of 20 equally likely events. What is the probability of each?
- When guessing randomly on a true/false question on a test, what is the probability that you get the answer correct?
- When guessing randomly on a multiple-choice question on a test that has 4 possible answers, what is the probability that you get the answer correct?

- 1
- 0
- 1/20 = 0.05
- 1/2 = 0.5
- 1/4 = 0.25

Based on a randomly drawn (and replaced) sample from a candy bin that consists of 428 green candies and 152 yellow candies, estimate the probability of drawing a green candy. Someone claims that when the bin was filled, three fourths of the candies were green. Do you believe this person?

$$\frac{428}{428+152} \approx 0.7379$$ The difference does not seem unusually large. Consequently, the sample does not seem to provide any evidence to refute the person's claim.

Construct the sample space for the 8 different (but equally likely) ways a couple could have 3 children, with regard to their gender (Male or Female). Based on this sample space and an assumption that the probability of having a boy and the probability of having a girl are both $0.5$, what is the probability of having all girls if a couple has 3 children? What is the probability of having at least one child of each gender?

Sample space: $$\begin{array}{cccccccc} GGG & GGB & GBG & GBB & BGG & BGB & BBG & BBB \end{array}$$ $P(\textrm{all girls}) = 1/8$

$P(\textrm{at least one of each gender}) = 6/8 = 3/4$Construct the sample space for the possible sequences of heads or tails that could result from four flips of a coin. Find the probability that exactly two of four flips are tails. Then find the probability all of the flips are heads.

Sample space: $$\begin{array}{cccccccc} HHHH & HHHT & HHTH & HHTT & HTHH & HTHT & HTTH & HTTT\\ THHH & THHT & THTH & THTT & TTHH & TTHT & TTTH & TTTT \end{array}$$ $P(\textrm{exactly 2 tails}) = 6/16 = 3/8$

$P(\textrm{all heads}) = 1/16$An electronic circuit is created to light a lamp. There are two buttons (named A and B) in the circuit, each of which can create a "short-to-ground" that will result in the lamp failing to light. List all of the possibilities for the state of the two buttons (pressed or not pressed) at any given moment. Assuming each of these are equally likely, what is the probability that the lamp lights in a given moment? What is the probability that the lamp fails to light?

Possibilities: $$\begin{array}{r|c|c} & \textrm{A is pressed} & \textrm{A is not pressed}\\\hline \textrm{B is pressed} & \textrm{fails to light} & \textrm{fails to light}\\\hline \textrm{B is not pressed} & \textrm{fails to light} & \textrm{lights}\\ \end{array}$$ $P(\textrm{lamp lights}) = 1/4$

$P(\textrm{lamp fails to light}) = 3/4$In rolling a single die, let $A$ represent rolling an even value and $B$ represent rolling a value greater than 4. Find $P(A \textrm{ or } B)$ in two ways: a) using the addition rule and b) considering the sample space.

There are 4 out of 6 values in the sample space that fit this criteria (i.e., $\{2,4,5,6\}$), so the probability is $4/6 = 2/3$.

Using the addition rule, we have: $P(\textrm{even}) + P(\gt4) - P(\textrm{even and } \gt 4) = 3/6 + 2/6 - 1/6 = 4/6 = 2/3$

In rolling a single die, let $A$ represent rolling an odd value and $B$ represent rolling a $6$. Find $P(A \textrm{ and } B)$ in two ways: a) using the multiplication rule and b) considering the sample space.

A and B can't both happen. A number can't be both odd and $6$. So the probability is $0$.A single card is drawn from a deck. Find the probability of selecting:

- a 4 or a diamond
- a club or a diamond
- a jack or a black card

$\displaystyle{\frac{4}{52} + \frac{13}{52} - \frac{1}{52} \doteq 0.3077}$

$\displaystyle{\frac{13}{52} + \frac{13}{52} = 0.5}$

$\displaystyle{\frac{4}{52} + \frac{26}{52} - \frac{2}{52} \doteq 0.5385}$

At a used-book sale, 100 books are adult books and 160 are children's books. Of the adult books, 70 are nonfiction while 60 of the children's books are nonfiction. If a book is selected at random, find the probability that it is:

- fiction
- not a children's nonfiction book
- an adult book or a children's nonfiction book

$\displaystyle{\frac{130}{260} = 0.5}$

$\displaystyle{1 - \frac{60}{260} \doteq 0.7692}$

$\displaystyle{\frac{100}{260} + \frac{60}{260} \doteq 0.6154}$

When two dice are rolled, find the probability of getting a sum that is

- 5 or 6
- greater than 9
- less than 4 or greater than 9
- divisible by 4

$\displaystyle{\frac{4+5}{36} = \frac{1}{4}}$

$\displaystyle{\frac{3+2+1}{36} = \frac{1}{6}}$

$\displaystyle{\frac{1+2+3+2+1}{36} = \frac{1}{4}}$

$\displaystyle{\frac{3+5+1}{36} = \frac{1}{4}}$

Suppose two coins are tossed, and you know that at least one of them resulted in "tails". What is the probability that they are both tails?

Sample space reduced to $\{HT,TH,TT\}$, so $\displaystyle{\frac{1}{3}}$In drawing two cards from a standard deck, what is the probability of drawing an ace on the first draw and a king on the second draw?

$\displaystyle{\frac{4}{52} \cdot \frac{4}{51} \doteq 0.0060}$Roll a single die. What is the probability that one rolls a 1 or 2 given that one rolled an even value?

$2$ of $6$ possibilities, so $\displaystyle{\frac{1}{3}}$At a large university, the probability that a student takes calculus and is on the dean's list is 0.042. The probability that a student is on the dean's list is 0.21. Find the probability that the student is taking calculus given that he or she is on the dean's list.

$\displaystyle{\frac{0.042}{0.21} = 0.2}$Eighty students in a school cafeteria were asked if they favored a ban on smoking in the cafeteria. The results of the survey are shown in the table below: $$\begin{array}{l|c|c|c} & \textrm{Favor} & \textrm{Oppose} & \textrm{No Opinion}\\\hline \textrm{Freshmen} & 15 & 27 & 8\\\hline \textrm{Sophomore} & 23 & 5 & 2\\\hline \end{array}$$ If a student is selected at random, find the probability that

- given the student is a freshman, he or she opposes the ban
- given the student favors the ban, he or she is a sophomore

$\displaystyle{\frac{27}{15+27+8} = 0.54}$

$\displaystyle{\frac{23}{15+23} \doteq 0.6053}$

Find the probability that if a coin is tossed twice, the first toss is "heads", while the second is "tails"

$\displaystyle{\frac{1}{2} \cdot \frac{1}{2} = 0.25}$If 37% of high school students said that they exercise regularly, find the probability that 5 randomly selected high school students will say that they exercise regularly.

$\displaystyle{(0.37)^5 \doteq 0.0069}$If 2 cards are selected from a standard deck of 52 cards without replacement, find the probability that

- both are spades
- both are the same suit

$\displaystyle{\frac{13}{52} \cdot \frac{12}{51} \doteq 0.0588}$

$\displaystyle{\frac{12}{51} \doteq 0.2353}$

The U.S. Department of Health and Human Services reports that 15% of Americans have chronic sinusitis. If 5 people are selected at random, find the probability that at least one has chronic sinusitis.

Use the complement. $\quad 1 - (0.85)(0.85)(0.85)(0.85)(0.85) = 1 - 0.85^5 \doteq 0.5563$A jar contains 8 red marbles, 9 blue marbles, and 10 green marbles. Four marbles are chosen at random without replacement. Find the probability of getting

- all green marbles
- 2 red and 2 blue marbles
- no green marbles
- exactly 2 green marbles
- at most 2 green marbles
- marbles that are all the same color

$\displaystyle{\frac{10}{27} \cdot \frac{9}{26} \cdot \frac{8}{25} \quad \textrm{or} \quad \frac{({}_{10} C_4)}{({}_{27} C_4)}}$

$\displaystyle{\frac{({}_8 C_2)({}_9 C_2)}{({}_{27} C_4)}}$

$\displaystyle{\frac{17}{27} \cdot \frac{16}{26} \cdot \frac{15}{25} \cdot \frac{14}{24} \quad \textrm{or} \quad \frac{({}_{17} C_4)}{({}_{27} C_4)}}$

$\displaystyle{\frac{({}_{10} C_2)({}_{17} C_2)}{({}_{27} C_4)}}$

$\displaystyle{\frac{({}_{10} C_0)({}_{17} C_4)}{({}_{27} C_4)} + \frac{({}_{10} C_1)({}_{17} C_3)}{({}_{27} C_4)} + \frac{({}_{10} C_2)({}_{17} C_2)}{({}_{27} C_4)}}$

$\displaystyle{\frac{({}_{8} C_4)}{({}_{27} C_4)} + \frac{({}_{9} C_4)}{({}_{27} C_4)} + \frac{({}_{10} C_4)}{({}_{27} C_4)}}$

What is the probability of getting at least one head in the tossing of a coin 3 times?

Use the complement: 1 - probability of getting all tails = 1 - 1/8 = 7/8Assume two fair dies are rolled. (a) What is the probability of rolling a sum of 7 or 11? (b) What is the probability that the sum is a prime number? (c) What is the probability that the sum is greater than 7 if you already know that the number showing on one die is 3? (d) What is the probability that the sum is at least 7? (e) What is the probability that the sum is even and is greater than 7?

Draw out the sample space to answer these questions. (a) P(7) + P(11) = 6/36 + 2/36 = 8/36 = 2/9; (b) Prime numbers: 2, 3, 5, 7, 11. There are 15 sums that are prime in the sample space, so 15/36; (c) There are 11 possibilities in the sample space where one die could be 3. Of these 11 possibilities, 4 sums are greater than 7. 4/11; (d) There are 21 possible ways to get a sum of 7 or 8 or 9 or 10 or 11 or 12. 21/36 = 7/12; (e) There are 9 possible ways to get a sum of 8 or 10 or 12. 9/36 = 1/4.Two people stranded on a tropical island each roll a standard die to determine who gets the last banana on the island. If the highest number rolled is a 1,2,3, or 4, player #1 wins. If the highest number rolled is a 5 or 6, player #2 wins. Who has the higher probability of winning?

Player #2 has the higher probability of 20/36. Watch the below video to see why:Four people are being considered for promotion from a pool of qualified applicants made up of 20 females and 15 males. The four selected were all males. What is the probability of this situation happening by chance?

The numerator is a count of the number of ways of getting 4 males from the 15 available (a combination) and getting no females from the 20 available (a combination) or $({}_{15}C_4)({}_{20}C_0)$, these two combinations are multiplied together (fundamental counting theorem). The denominator is a count of the number of ways of getting 4 people from the 35 available (a combination) or ${}_{35}C_4$. This fraction reduces to be 39/1496 or approximately 0.026, highly unlikely.If it is known that 80% of all freshmen have had at least one drink, what is the probability of the following if 5 freshmen are selected at random: (a) all 5 freshmen have had at least one drink; at least one freshman has not had at least one drink?

(a) $(0.80)^5 \approx 0.328$; (b) $1 - 0.328 = 0.672$ (use of complement)A jury of 12 is to be selected from a list of 35 mailes and 15 females. What is the probability that there is no more than one female on the jury?

Two terms are added together. The denominator for each fraction is ${}_{50}C_{12}$. For $0$ females and all males, the numerator is $({}_{15}C_0)({}_{35}C_{12})$. For exactly one female and eleven males, the numerator is $({}_{15}C_1)({}_{35}C_{11})$. The approximate value of the two terms is $0.00687 + 0.05155 = 0.05842$.You are in a room of 12 people. What is the probability that at least two of these people will have the same birthday?

Birthday problem (see #39) with $n=12$. $1 - 0.833 = 0.167$ (approximately)

A box contains 8 good light bulbs and 4 that are defective. If three bulbs are randomly selected from the box without replacement, find the probability that at least one will be defective.

P(at least one defective) = 1 - P(all are good), which equals $$1 - \frac{({}_8C_3)({}_4C_0)}{({}_{12}C_3)} \approx 0.745$$Five people are to be randomly selected for a committee from a class consisting of 10 freshmen and 8 sophomores. What is the probability that all five of those selected are sophomores.

$$\frac{({}_8C_5)({}_{10}C_0)}{({}_{18}C_5)} \approx 0.007$$On a multiple choice test with answers a, b, c, and d, what is the probability of getting the first three questions correct (assuming you know nothing about the questions)?

Independent events and multiplication rule, $1/64$There are 3 defective calculators in a box of 100. Out of five selected, what is the probability that: (a) four are good and one is defective; (b) there are at most two defective; (c) all five are good

Counting with probability: (a) $0.138$; (b) $0.999938$, just about $1$; (c) $0.856$Six red balls and four blue balls are placed in an urn. Four balls are drawn at random from the urn. Find the probability that all four are blue if (a) each ball is returned to the urn before the next is drawn, (b) the balls are not returned to the urn.

The first is independent and the second is dependent (on what was picked first): (a) $16/625 = 0.0256$; (b) $1/210$A committee of six is randomly selected from a club with 18 male and 12 female members. (a) What is the probability that there is at least one female on the committee? (b) Find the probability that there are three males and three females on the committee.

(a) Use the complement, 1 - P(no female) = $1 - 0.0313 = 0.9687$, approximately; (b) Counting with probability: $0.3023$Toss a pair of tetrahedron dice, each numbered 1, 2, 3, 4. Let X be the sum of the numbers. Find the probability of getting a sum that is: (a) at least 6; (b) at most 4; (c) more than 6; (d) less than 4; (e) and even number; (f) a prime number; (g) divisible by 4.

Draw out the sample space... (a) 3/8; (b) 3/8; (c) 3/16; (d) 3/16; (e) 1/2; (f) 9/16; (g) 1/4-
In a certain lottery, one must write down a sequence of 5 different numbers, each between 1 and 49, on a slip of paper. From an air-hopper containing forty-nine balls, numbered $1$ to $49$, a sequence of $5$ balls are drawn without replacement. To win one of the prizes in the lottery, the numbers on one's slip must match the numbers on the drawn balls, although their order doesn't matter. Find the probability of winning the aforementioned prize in two ways: first using permutations, and then using combinations.

There are ${}_{49} P_5$ possible ways a sequence of $5$ numbered balls could be drawn. For any of these, there are 5! winning slips, resulting in the probability of winning being $$\begin{array}{rcl} \displaystyle{\frac{5!}{{}_{49} P_5}} &=& \displaystyle{\frac{\quad \quad 5! \quad \quad}{\displaystyle{\frac{49!}{(49-5)!}}}}\\\\ &=& \displaystyle{\frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{49 \cdot 48 \cdot 47 \cdot 46 \cdot 45}}\\\\ &=& \displaystyle{\frac{1}{1906884}} \end{array}$$

Alternatively (

*and more simply*), there are ${}_{49} C_5$ ways a set of $5$ numbered balls can be drawn, not counting different orderings of the same balls as different. For any of these, there is only one set of numbers that could be written on a winning slip, resulting in the probability of winning being $$\frac{1}{{}_{49} C_5} = \frac{1}{1906884}$$ Letters are written to $8$ people and mailed one after another. Assuming that each person has a different last name, what is the probability that the letters are mailed in alphabetical order by last name?

$$\frac{1}{8!} = \frac{1}{40320}$$There are $14$ cities in Georgia that you would like to tour. You plan a vacation to visit $5$ of them. If the cities are selected at random, what is the probability that your vacation involves touring Macon, Valdosta, Savannah, Augusta, and Atlanta, in that order?

$$\frac{1}{{}_{14} P_5} = \frac{1}{240240}$$**The Birthday Problem.**Find the probability that 2 or more people in a group of $n$ people have the same birthday.Use the complement: $$\begin{array}{rcl} P(\textrm{2 or more share a b-day}) &=& 1 - P(\textrm{all different b-days})\\\\ &=& \displaystyle{1 - \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \cdots \frac{365-(n-1)}{365}}\\\\ &=& \displaystyle{1 - \frac{(365)(364)(363)\cdots(365-(n-1))}{365^n}}\\\\ &=& \displaystyle{1 - \frac{{}_{365}P_n}{365^n}} \end{array}$$

Note the last step uses permutations only to shorten the formula -- there are no actual "orderings" being counted here.

Watch this video to see the calculations above explained!

In the game of Craps, a player rolls a pair of dice. If, on this first roll, the sum of the dice is 7 or 11, they win immediately. If the sum of the dice is 2, 3, or 12, they lose immediately. If instead, they roll some other value $x$ (called the "point"), they continue to roll the dice until either the point is rolled (where the player then wins), or a 7 is rolled (where the player then loses). What is the probability that the player wins a game of Craps?

$$\begin{array}{rcl} P(win) &=& P(\textrm{win with 7 or 11}) + P(\textrm{win},pt=4) + P(\textrm{win},pt=5) + P(\textrm{win},pt=6) +\\ && P(\textrm{win},pt=8) + P(\textrm{win},pt=9) + P(\textrm{win},pt=10)\\ &=& \frac{8}{36} + P(4 \textrm{ on 1st})P(4 \textrm{ before } 7) + P(5 \textrm{ on 1st})P(5 \textrm{ before } 7) + \cdots\\ &=& \frac{8}{36} + \frac{3}{36} \cdot \frac{3}{9} + \frac{4}{36} \cdot \frac{4}{10} + \frac{5}{36} \cdot \frac{5}{11} + \frac{5}{36} \cdot \frac{5}{11} + \frac{4}{36} \cdot \frac{4}{10} + \frac{3}{36} \cdot \frac{3}{9}\\ &\doteq& 0.4929 \end{array}$$In rolling a pair of dice, what is the probability that you roll

- at least a sum of 7?
- at most a 7 given that at least one die shows a 3

- Look at sample space of $36$ possibilities. $21/36$
- Sample space reduced to $11$ possible rolls, $7$ of which are at most $7$, so $7/11$

Toss a coin 3 times. What is the probability that

- there is at most one head
- one gets 2 heads given that the first toss is a tail

- 4/8 = 1/2
- Sample space reduced to 4 possibilities. probability is 1/4

For a bridge hand of 13 cards, find the probability that you have

- exactly 9 spades
- all 4 aces and no face cards
- 5 spades, 6 hearts, and 2 diamonds, given that none of your cards are aces

- $\displaystyle{\frac{({}_{13}C_{9})({}_{39}C_{4})}{({}_{52}C_{13})} \doteq 0.0000926}$
- $\displaystyle{\frac{({}_{36}C_{9})}{({}_{52}C_{13})} \doteq 0.000148}$
- $\displaystyle{\frac{({}_{12}C_{5})({}_{12}C_{6})({}_{12}C_{2})}{({}_{48}C_{13})} \doteq 0.0002503}$

A 6-sided die is rolled 4 times. Find the probability of rolling at least one 6.

Use the complement. $1 - P(0) = 1 - (5/6)^4 \doteq 0.5177$Two 6-sided dice are rolled 24 times. Find the probability of rolling at least one sum of 12 (i.e., two sixes).

$P(12 \textrm{ on one roll}) = 1/36$. So,

$\displaystyle{\begin{array}{rcl} P(\textrm{at least one twelve}) &=& 1 - P(\textrm{no twelves})\\ &=& 1 - (35/36)^{24}\\ &\doteq& 0.4914 \end{array}}$