Probability

Probability

Probability of a simple event \(A\) is:

\(P(A)=\frac{number~of~favourable ~outcomes}{number~of~all~possible~outcomes}\)

Simple events

We roll a standard fair playing die. Calculate the probabilities of the following events:

A:   The number shown is a 6.

B:   The number shown is greater than 4.

C:   The number shown is less than 4.
Solutions:    \(P(A)=\frac{1}{6}\approx16.7\%\);     \(P(B)=\frac{2}{6}=\frac{1}{3}\approx33.3\%\);     \(P(C)=\frac{3}{6}=\frac{1}{2}=50\%\)
We roll a fair twenty-sided die. Calculate the probabilities of the following events:

A:   The number shown is odd.

B:   The number shown is a prime number.

C:   The number shown is a solution of the equation \(x^2-8x+15=0\).
Solutions:    \(P(A)=\frac{10}{20}=\frac{1}{2}=50\%\);     \(P(B)=\frac{8}{20}=\frac{2}{5}=40\%\);     \(P(C)=\frac{2}{20}=\frac{1}{10}=10\%\)
Two standard fair playing dice are rolled. Calculate the probabilities of the following events:

A:   The numbers shown are 1 and 1.

B:   The sum of the numbers shown is 3.

C:   The sum of the numbers shown is 4.

D:   The sum of the numbers shown is 5.

E:   The numbers shown are equal.
Solutions:    \(P(A)=\frac{1}{36}\);     \(P(B)=\frac{2}{36}=\frac{1}{18}\);     \(P(C)=\frac{3}{36}=\frac{1}{12}\);     \(P(D)=\frac{4}{36}=\frac{1}{9}\);     \(P(E)=\frac{6}{36}=\frac{1}{6}\)
Three playing dice are rolled. Calculate the probabilities of the following events:

A:   The sum of the numbers shown is 3.

B:   The sum of the numbers shown is 4.

C:   All three numbers shown are equal.
Solutions:    \(P(A)=\frac{1}{216}\);     \(P(B)=\frac{3}{216}=\frac{1}{72}\);     \(P(C)=\frac{6}{216}=\frac{1}{36}\)
A standard deck of 52 playing cards consists of four suits: the hearts (♥), the diamonds (♦), the spades (♠) and the clubs (♣). Each suit includes thirteen cards labelled: A (ace or 1), 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen) and K (king). We pick one card at random. Find the probabilities of the following events:

A:   The card is an ace.

B:   The card is a spade.

C:   The card is the queen of hearts.

D:   The card is not a king.
Solutions:    \(P(A)=\frac{4}{52}=\frac{1}{13}\);     \(P(B)=\frac{13}{52}=\frac{1}{4}\);     \(P(C)=\frac{1}{52}\);     \(P(D)=\frac{48}{52}=\frac{12}{13}\)

Complement, union and intersection of events

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Complementary event:

\(P(A')=1-P(A)\)
A wheel of fortune is divided in twelve sectors numbered with numbers 1 to 12. We spin the wheel and get a number. Find the probabilities of the following events:

A:   The number is a multiple of 5.

B:   The number is not a multiple of 5.
Solutions:    \(P(A)=\frac{2}{12}=\frac{1}{6}\);     \(P(B)=P(A')=1-P(A)=\frac{10}{12}=\frac{5}{6}\)
We pick (at random) one card out of the standard deck of 52 playing cards. Find the probabilities of the following events:

A:   The card is an ace or a king.

B:   The card is a spade or a club.

C:   The card is a queen or a heart.

D:   The card is neither a queen nor a heart.
Solutions:    \(P(A)=\frac{8}{52}=\frac{2}{13}\);     \(P(B)=\frac{26}{52}=\frac{1}{2}\);     \(P(C)=\frac{16}{52}=\frac{4}{13}\);     \(P(D)=\frac{36}{52}=\frac{9}{13}\)
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Union of events:

\(P(A\cup B)= P(A)+P(B)-P(A\cap B)\)
There are 20 students in a class. They can learn Russian or Chinese (or both). Each student must learn at least one of these two languages. There are 15 students learning Russian and 9 students learning Chinese. We choose a random student. Find the probabilities of the following events:

A:   This student learns Russian.

B:   This student learns Chinese.

C:   This student learns both: Russian and Chinese.

D:   This student learns exactly one of these two languages.
Solutions:    \(P(A)=\frac{15}{20}=75\%\);     \(P(B)=\frac{9}{20}=45\%\);     \(P(C)=\frac{4}{20}=20\%\);     \(P(D)=\frac{16}{20}=80\%\)
There are 25 students in a class, 16 of them play football and 10 of them play basketball. 7 of them play both: football and basketball. One of the students is selected at random. Find the probabilities of the following events:

A:   This student plays football.

B:   This student plays football or basketball.

C:   This student plays basketball but not football.

D:   This student plays exactly one of these two sports.

E:   This student plays neither football nor basketball.
Solutions:    \(P(A)=\frac{16}{25}=64\%\);     \(P(B)=\frac{19}{25}=76\%\);     \(P(C)=\frac{3}{25}=12\%\);     \(P(D)=\frac{12}{25}=48\%\);     \(P(E)=\frac{6}{25}=24\%\)
A small town has 6000 inhabitants, 3300 of them read the Morning Mirror and 2100 of them read the Evening Events. There are 1500 inhabitants who don't read any of these two magazines. We are going to interview an inhabitant selected at random. Find the probabilities of the following events:

A:   This inhabitant reads both magazines.

B:   This inhabitant reads at least one of these two magazines.
Solutions:    \(P(A)=\frac{900}{6000}=15\%\);     \(P(B)=\frac{4500}{6000}=75\%\)
A jar contains 8 blue marbles, 7 red marbles and 5 green marbles. Two marbles are extracted in a row. The first marble is replaced before extracting the second one. Find the probabilities of the following events:

A:   Both marbles are blue.

B:   Both marbles are red.

C:   Both marbles are green.
Solutions:    \(P(A)=\frac{8}{20}\cdot\frac{8}{20}=16\%\);     \(P(B)=\frac{7}{20}\cdot\frac{7}{20}=12.25\%\);     \(P(C)=\frac{5}{20}\cdot\frac{5}{20}=6.25\%\)
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Product of independent events:

\(P(A\,B)=P(A)\cdot P(B)\)

Product of dependent events:

\(P(A\,B)=P(A)\cdot P(B\,|A)\)
A jar contains 8 blue marbles, 7 red marbles and 5 green marbles. Two marbles are extracted in a row. The first marble is not replaced. Find the probabilities of the following events:

A:   Both marbles are blue.

B:   Both marbles are red.

C:   Both marbles are green.
Solutions:    \(P(A)=\frac{8}{20}\cdot\frac{7}{19}\approx14.7\%\);     \(P(B)=\frac{7}{20}\cdot\frac{6}{19}\approx11.1\%\);     \(P(C)=\frac{5}{20}\cdot\frac{4}{19}\approx5.26\%\)
A jar contains 5 white and 7 black marbles. You draw 3 marbles in a row (at random). Find the probabilities of choosing 3 black marbles:

(a)   if you replace each marble before extracting the next one,

(b)   if you don't replace the marbles.
Solutions:    (a)  \(P(A)=\frac{7}{12}\cdot\frac{7}{12}\cdot\frac{7}{12}=\frac{343}{1728}\approx19.8\%\);     (b)  \(P(B)=\frac{7}{12}\cdot\frac{6}{11}\cdot\frac{5}{10}=\frac{7}{44}\approx15.9\%\)
A jar contains 4 red, 5 blue and 7 yellow marbles. You draw 2 marbles in a row (without replacing them). Find the probabilities of the following events:

A:   Both marbles are yellow.

B:   The first is yellow and the second is blue.

C:   At least one of the marbles is yellow.
Solutions:    \(P(A)=\frac{7}{16}\cdot\frac{6}{15}=\frac{7}{40}=17.5\%\);     \(P(B)=\frac{7}{16}\cdot\frac{5}{15}=\frac{7}{48}\approx14.6\%\);     \(P(C)=1-\frac{9}{16}\cdot\frac{8}{15}=\frac{7}{10}=70\%\)
A wheel of fortune is divided in 36 sectors numbered from 1 to 36. We spin the wheel 3 times, getting one number at a time. Find the probabilities of the following events:

A:   All three numbers are even.

B:   The first number is less than 10 and the other two numbers are greater than 10.

C:   The first number is a multiple of 2, the second is a multiple of 3 and the third is a multiple of 4.

D:   None of the numbers is a multiple of 9.
Solutions:    \(P(A)=\frac{1}{8}=12.5\%\);     \(P(B)=\frac{169}{1296}\approx13.0\%\);     \(P(C)=\frac{1}{24}\approx4.17\%\);     \(P(D)=\frac{512}{729}\approx70.2\%\)
There are 25 pupils in a class: 10 boys and 15 girls. Three representatives are chosen at random. Find the probabilities of the following events:

A:   All three representatives are girls.

B:   At least one of the representatives is a boy.
Solutions:    \(P(A)=\frac{15}{25}\cdot\frac{14}{24}\cdot\frac{13}{23}=\frac{91}{460}\approx19.8\%\);     \(P(B)=1-P(A)=\frac{369}{460}\approx80.2\%\)
Two cards are chosen at random out of the standard deck of 52 playing cards. Find the probabilities of the following events:

A:   Both card are hearts.

B:   Both cards are aces.
Solutions:    \(P(A)=\frac{13}{52}\cdot\frac{12}{51}=\frac{1}{17}\approx5.88\%\);     \(P(B)=\frac{4}{52}\cdot\frac{3}{51}=\frac{1}{221}\approx0.452\%\)
Three cards are chosen at random out of the standard deck of 52 playing cards. Find the probabilities of the following events:

A:   Exactly one of the cards is an ace.

B:   At least one of the cards is an ace.
Solutions:    \(P(A)=\frac{4}{52}\cdot\frac{48}{51}\cdot\frac{47}{50}+\frac{48}{52}\cdot\frac{4}{51}\cdot\frac{47}{50}+\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{4}{50} \approx20.4\%\);     \(P(B)=1-\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{46}{50}\approx21.7\%\)

When multiple objects are extracted at the same time you can use the binomial coefficient \({n \choose r}\) to find the probability. Binomial coefficient \({n \choose r}\) tells us in how many ways \(r\) objects can be selected out of \(n\) objects available. This function is usually labeled nCr on calculators.

There are 12 girls and 8 boys in a class. They select four random representatives. Find the probabilities of the following events:

A:   All representatives are girls.

B:   All representatives are boys.
Solutions:    \(P(A)=\frac{12\choose 4}{20\choose 4}\approx10.2\%\);     \(P(B)=\frac{8\choose 4}{20\choose 4}\approx1.44\%\)
Five cards are chosen at random out of the standard deck of 52 playing cards. Find the probabilities of the following events:

A:   All five cards are non-aces.

B:   At least one of the cards is an ace.
Solutions:    \(P(A)=\frac{48\choose 5}{52\choose 5}\approx65.9\%\);     \(P(B)=1-P(A)\approx34.1\%\)

Conditional probability

There are 500 students who attend the Sophie Germain Middle School, 225 of them are male and 275 female. Among males 45 like prime numbers and among females 88 like prime numbers. Calculate the following probabilities:

(a)   We choose an arbitrary student. Find the probability that this student likes prime numbers.

(b)   We choose a male student. Find the probability that he likes prime numbers.

(c)   We choose a female student. Find the probability that she likes prime numbers.

(d)   We choose a student who likes prime numbers. Find the probability that this student is a female.
Solutions:    (a)  \(P(A)=\frac{133}{500}=26.6\%\);     (b)  \(P(A|M)=\frac{45}{225}=20\%\);     (c)  \(P(A|F)=\frac{88}{275}=32\%\);     (d)  \(P(F|A)=\frac{88}{133}\approx66.2\%\)
In our country 20 000 people died last year, 8 000 of them were smokers and 12 000 non-smokers. Among smokers 5 920 died of lung cancer. Among non-smokers 480 died of lung cancer. Calculate the following probabilities:

(a)   Find the probability that a person who died last year was a smoker.

(b)   Find the probability that a person who died last year was a non-smoker.

(c)   Find the probability that a person who died last year had lung cancer.

(d)   Find the probability that a smoker who died last year had lung cancer.

(e)   Find the probability that a person who died of lung cancer was a smoker.
Solutions:    (a)  \(P(S)=40\%\);     (b)  \(P(N)=60\%\);     (c)  \(P(L)=32\%\);     (d)  \(P(L|S)=74\%\);     (e)  \(P(S|L)=92.5\%\)
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Events are independent if and only if:

\(P(A\cap B)=P(A)\,P(B)\)
The wheel of fortune is divided in 9 sectors labeled with numbers 1 to 9. We spin this wheel two times.

(a)   Find the probabilities of the following events:

A: the first number is even,

B: the second number is even.

(b)   Show that events A and B are independent.
Solutions:    (a)  \(P(A)=\frac{4}{9},~ P(B)=\frac{4}{9}\);     (b)  \(P(A\cap B)=\frac{16}{81}=P(A)\cdot P(B)\), so the events are independent
There are 6 white and 4 black marbles in a jar. We extract two marbles in a row (without returning the first one). Find the probabilities of the following events:

(a)   Find the probabilities of the following events:

A: the first marble is black,

B: the second marble is black.

(b)   Verify if events A and B are independent.
Solutions:    (a)  \(P(A)=\frac{4}{10},~ P(B)=\frac{4}{10}\);     (b)  \(P(A\cap B)=\frac{2}{15}\ne P(A)\cdot P(B)\), so the events are not independent
A small town has 5000 inhabitants. They have two newspapers: 1500 inhabitants read the Morning Magazine and 2000 inhabitants read the Evening Events. There are 100 inhabitants who read both newspapers.

(a)   Find the probability of A, that a randomly chosen person reads the Morning Magazine.

(b)   Find the probability of B, that a randomly chosen person reads the Evening events.

(c)   Find the probability that a randomly chosen person reads both newspapers.

(d)   Find the probability that a randomly chosen reader of the Evening Events reads the Morning Magazine, too.

(e)   Verify if events A and B are independent.
Solutions:    (a)  \(P(A)=30\%\);     (b)  \(P(B)=40\%\);     (c)  \(P(A\cap B)=2\%\);     (d)  \(P(A|B)=5\%\);     (e)  \(P(A\cap B)\ne P(A)\cdot P(B)\), so the events are not independent
There are 20 students in a class and they can play two sports: 5 students play basketball and 8 students play football. There are 2 students who play basketball and football. We choose a random student and we study the following events:

B: this student plays basketball,

F: this student plays football.

(a)   Calculate \(P(B),~ P(F)\) and \(P(B\cap F)\).

(b)   Verify if events B and F are independent.
Solutions:    (a)  \(P(B)=\frac{1}{4},~ P(F)=\frac{2}{5},~ P(B\cap F)=\frac{1}{10}=P(B)\cdot P(F)\);     (b)  the events are independent

Total probability and Bayes' theorem

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Total probability:

\(P(A)=P(H_1)P(A|H_1)+P(H_2)P(A|H_2)\)
The first jar contains 1 white and 3 black marbles. The second jar contains 3 white and 4 black marbles. Firstly we select (at random) a marble from the first jar and put it in the second jar. Then we select (at random) a marble from the second jar. Find the probability that the second marble is black.
Solutions: \(P(A)=\frac{19}{32}\approx59.4\%\)
The first jar contains 3 red and 2 blue marbles. The second jar contains 1 red and 4 blue marbles. Firstly we select a random marble from the first jar and put it in the second jar. Then we select a random marble from the second jar. Find the probability that the second marble is red.
Solutions: \(P(A)=\frac{4}{15}\approx26.7\%\)
An international corporation has factories all around the world. Electric scooters are produced in two factories: 70% of scooters are produced in Western Europe and 30% are produced in Transnistria. Among scooters produced in Western Europe 1% are defective. Among scooters produced in Transnistria 16% are defective. You buy a scooter and you don't know where it was produced. Find the probability that it is defective.
Solutions: \(P(A)=5.5\%\)
We have two jars, each of them containing 10 marbles. There are 6 white and 4 black marbles in the first jar and 2 white and 8 black in the second jar. First we select (at random) one of the jars. Then we pick (at random) one marble from the selected jar.

(a)   Draw the corresponding tree diagram.

(b)   Find the probability that the chosen marble is black.
Solutions:    (b)  \(P(A)=\frac{3}{5}=60\%\)
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Bayes' theorem:

\({\displaystyle P(H_1|A)=\frac{P(H_1)P(A|H_1)}{P(A)}}\)
We have two jars, each of them containing 10 marbles. There are 6 white and 4 black marbles in the first jar and 2 white and 8 black in the second jar. First we select (at random) one of the jars. Then we pick (at random) one marble from the selected jar. Then we take a look at the chosen marble and we see that it is black.

(a)   Find the probability that this black marble was picked from the first jar.

(b)   Find the probability that this black marble was picked from the second jar.
Solutions:    (a)  \(P(H_1|A)=\frac{1}{3}\approx33.3\%\);     (b)  \(P(H_2|A)=\frac{2}{3}\approx66.7\%\)
The first jar contains 1 red and 3 blue marbles. The second jar contains 2 red and 7 blue marbles. You first pick a marble from the first jar and you put it in the second jar (without looking at it). Then you pick a marble from the second jar.

(a)   Find the probability that the second marble is red.

(b)   Knowing that the second marble is red, find the probability that the first marble was red too.
Solutions:    (a)  \(P(A)=\frac{9}{40}=22.5\%\);     (b)  \(P(H_1|A)=\frac{1}{3}\approx33.3\%\)
Statistics shows that 2% od the inhabitants of a certain country are infected with the MIV virus. There's a quick test for discovering the infection, but it's only 99% reliable (if a person is infected the test is positive in 99% cases and if a person is not infected the test is negative in 99% cases). We are considering a randomly chosen person with the positive test result. Calculate the probability that this person is infected.
Solutions: \(P(H_1|A)\approx66.9\%\)

Repeated trials

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Binomial probability (probability that \(A\) happens \(r\) times in \(n\) trials):

\(P_r (A)={n\choose r}\, p^r \, \left(1-p\right)^{n-r}\)
You roll a standard playing die 5 times.

(a)   Find the probability of getting a 6 exactly once.

(b)   Find the probability of getting a 6 exactly twice.

(c)   Find the probability of getting a 6 exactly three times
Solutions:    (a)  \(P(A)\approx40.2\%\);     (b)  \(P(B)\approx16.1\%\);     (c)  \(P(C)\approx3.22\%\)
You toss a coin 4 times. Each time the coin can land in two different ways: one of the outcomes is called head and the other tail.

(a)   Find the probability of getting four tails.

(b)   Find the probability of getting at least one head.

(c)   Find the probability of getting exactly one head.

(d)   Find the probability of getting exactly two heads.
Solutions:    (a)  \(P(A)=\left(\frac{1}{2}\right)^4=\frac{1}{16}=6.25\%\);     (b)  \(P(B)=1-P(A)=\frac{15}{16}=93.75\%\);     (c)  \(P(C)=\frac{4}{16}=\frac{1}{4}=25\%\);     (d)  \(P(D)=\frac{6}{16}=\frac{3}{8}=37.5\%\)
Joe is a mediocre basketball player. His probability of making a basket in a single throw is 60%. Then he tries to do it 10 times

(a)   Find his probability of making exactly two baskets.

(b)   Find his probability of making exactly five baskets.

(c)   Find his probability of making exactly six baskets.
Solutions:    (a)  \(P(A)\approx 1.06 \%\);     (b)  \(P(B)\approx 20.1\%\);     (c)  \(P(C)\approx 25.1\%\)

Index