Probability

Simple events

1. 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:    \(P(A)=\frac{1}{6}\approx16.7\%\);     \(P(B)=\frac{2}{6}=\frac{1}{3}\approx33.3\%\);     \(P(C)=\frac{3}{6}=\frac{1}{2}=50\%\)
2. 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:    \(P(A)=\frac{10}{20}=\frac{1}{2}=50\%\);     \(P(B)=\frac{8}{20}=\frac{2}{5}=40\%\);     \(P(A)=\frac{2}{20}=\frac{1}{10}=10\%\)
3. 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:    \(P(A)=\frac{1}{36}\);     \(P(B)=\frac{2}{36}=\frac{1}{18}\);     \(P(C)=\frac{3}{36}=\frac{1}{12}\);     \(P(D)=\frac{4}{36}=\frac{1}{9}\);     \(P(E)=\frac{6}{36}=\frac{1}{6}\)
4. 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:    \(P(A)=\frac{1}{216}\);     \(P(B)=\frac{3}{216}=\frac{1}{72}\);     \(P(C)=\frac{6}{216}=\frac{1}{36}\)
5. 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

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:    \(P(A)=\frac{8}{52}=\frac{2}{13}\);     \(P(B)=\frac{26}{52}=\frac{1}{2}\);     \(P(C)=\frac{16}{52}=\frac{4}{13}\);     \(P(D)=\frac{36}{52}=\frac{9}{13}\)
7. 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:    \(P(A)=\frac{{13\choose 2}}{{52\choose 2}}=\frac{1}{17}\);     \(P(B)=\frac{{4\choose 2}}{{52\choose 2}}=\frac{1}{221}\)
8. A jar contains 8 blue marbles, 7 red marbles and 5 green marbles. Three marbles are extracted (at the same time). Find the probabilities of the following events:
   

   A:   All three marbles are blue.

   B:   All three marbles are red.

   C:   There are no green marbles.

   D:   There is at least one green marble.

   Solutions:    \(P(A)=\frac{14}{285}\approx4.91\%\);     \(P(B)=\frac{7}{228}\approx3.07\%\);     \(P(C)=\frac{91}{228}\approx39.9\%\);     \(P(D)=1-P(C)=\frac{137}{228}\approx60.1\%\)
9. 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:    \(P(A)=\frac{{15\choose 3}}{{25\choose 3}}=\frac{91}{460}\approx19.8\%\);     \(P(B)=1-P(A)=1-\frac{369}{460}\approx80.2\%\)
10. 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:    \(P(A)=\frac{16}{25}=64\%\);     \(P(B)=\frac{19}{25}=76\%\);     \(P(C)=\frac{3}{25}=12\%\);     \(P(D)=\frac{12}{25}=48\%\);     \(P(E)=\frac{6}{25}=24\%\)
11. 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:    \(P(A)=\frac{900}{6000}=15\%\);     \(P(B)=\frac{4500}{6000}=75\%\)
12. 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 pulling 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)  \(P(B)=\frac{7}{12}\cdot\frac{6}{11}\cdot\frac{5}{10}=\frac{7}{44}\approx15.9\%\)
13. 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:    \(P(A)=\frac{7}{16}\cdot\frac{6}{15}=\frac{7}{40}=17.5\%\);     \(P(B)=\frac{7}{16}\cdot\frac{5}{15}=\frac{7}{48}\approx14.6\%\);     \(P(C)=1-\frac{9}{16}\cdot\frac{8}{15}=\frac{7}{10}=70\%\)
14. 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:    \(P(A)=\frac{1}{8}=12.5\%\);     \(P(B)=\frac{169}{1296}\approx13.0\%\);     \(P(C)=\frac{1}{24}\approx4.17\%\);     \(P(D)=\frac{63}{64}\approx98.4\%\)

Conditional probability

15. 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)  \(P(F|A)=\frac{88}{133}\approx66.2\%\)
16. 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)  \(P(S|L)=92.5\%\)
17. 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)  \(P(A)=\frac{3}{5}=60\%\)
18. 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)  \(P(H_2|A)=\frac{2}{3}\approx66.7\%\)
19. 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)  \(P(H_1|A)=\frac{1}{3}\approx33.3\%\)
20. 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:    \(P(H_1|A)\approx66.9\%\)
21. 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)  \(P(D)=\frac{6}{16}=\frac{3}{8}=37.5\%\)
22. 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)  \(P(C)\approx3.22\%\)

Random variables

23. A wheel of fortune is divided in 15 sections numbered with values 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5. The random variable \(X\) is the number selected when the wheel is spun. Write the table of all possible results and their respective probabilities.
    Solutions:    \(\begin{array}{|l|c|c|c|c|c|}\hline \mathrm{Result}~~ x & 1 & 2 & 3 & 4 & 5 \cr\hline \mathrm{Probability}~~ P(X=x) & \frac{1}{3} & \frac{4}{15} & \frac{1}{5} & \frac{2}{15} & \frac{1}{15} \cr\hline \end{array}\)
24. Three marbles are extracted (without replacement) out of a jar which includes 3 white and 4 red marbles. The random variable \(X\) is the number of red marbles obtained. Write the table of probability distribution.
    Solutions:    \(\begin{array}{|l|c|c|c|c|}\hline x & 0 & 1 & 2 & 3 \cr\hline P(X=x) & \frac{1}{35} & \frac{12}{35} & \frac{18}{35} & \frac{4}{35} \cr\hline \end{array}\)
25. The random variable \(X\) is the number of heads obtained when you toss a coin four times.
    

    (a)   Write the table of probability distribution.

    (b)   Find the expected value \(E(X)\).

    Solutions:    (a)   \(\begin{array}{|l|c|c|c|c|c|}\hline x & 0 & 1 & 2 & 3 & 4 \cr\hline P(X=x) & \frac{1}{16} & \frac{1}{4} & \frac{3}{8} & \frac{1}{4} & \frac{1}{16} \cr\hline \end{array}\)     (b)  \(E(X)=2\)
26. The test is composed of three exercises: the first is on differentiation, the second on integration and the third one on probability. An average student can solve 65% of exercises on differentiation, 70% of exercises on integration and 80% of exercises on probability. The random variable \(X\) represents the number of exercises (0, 1, 2, or all 3) solved by an average student.
    

    (a)   Write the table of probability distribution.

    (b)   Find the expected value \(E(X)\).

    Solutions:    (a)   \(\begin{array}{|l|c|c|c|c|}\hline x & 0 & 1 & 2 & 3 \cr\hline P(X=x) & 2.1\% & 17.2\% & 44.3\% & 36.4\% \cr\hline \end{array}\)     (b)  \(E(X)=2.15\)
27. The table shows the probability distribution of a random variable \(X\).
    \(\begin{array}{|l|c|c|c|c|c|}\hline x & 2 & 4 & 6 & 8 & 10 & 12 \cr\hline P(X=x) & \frac{5}{20} & \frac{7}{20} & \frac{3}{20} & \frac{3}{20} & p & \frac{1}{20} \cr\hline \end{array}\)
    

    (a)   Find the unknown \(p\).

    (b)   Find the expected value \(E(X)\).

    Solutions:    (a)  \(p=\frac{1}{20}\);    (b)  \(E(X)=5.1\)
28. A standard fair playing die is rolled 10 times. The random variable \(X\) represents the number of sixes. Calculate:
    

    (a)   \(P(X=0)\).

    (b)   \(P(X=1)\).

    (c)   \(P(X\geqslant1)\).

    (d)   \(P(X\geqslant2)\).

    Solutions:    (a)  \(P(X=0)\approx16.2\%\);    (b)  \(P(X=1)\approx32.3\%\);    (c)  \(P(X\geqslant1)\approx83.8\%\);    (d)  \(P(X\geqslant2)\approx51.5\%\)
29. A standard fair playing die is rolled 120 times. The random variable \(X\) represents the number of sixes. Calculate \(E(X)\).
    Solutions:    \(E(X)=20\)      (Hint: This is the binomial distribution so you can use the formula \(E(X)=np\).)
30. The random variable \(X\) is distributed as \(X\sim B(60,0.2)\). Find \(E(X)\).
    Solutions:    \(E(X)=12\)

 Index