Kuulchat - Senior High School (S.H.S) Mathematics Probability Practice Test Questions

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PROBABILITY

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A number is chosen at random from the set {13, 14, ...., 30}. What is the probability that it is a prime number?

$\frac{5}{16}$

$\frac{1}{3}$

$\frac{5}{18}$

$\frac{3}{8}$

Set = {13, 14, ...., 30}

Set = {13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}

Number of samples = 18

Prime numbers are numbers with only two factors (1 and itself). Thus only 1 and the number itself can divide the number without a remainder.

Set of prime numbers from the set = {13, 17, 19, 23, 29}

Number of prime numbers = 5

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

P(Prime) = $\frac{5}{18}$

The probability that John and James pass an examination are $\frac{3}{4}$ and $\frac{3}{5}$ respectively.

Find the probability that both will fail.

$\frac{1}{10}$

$\frac{11}{20}$

$\frac{9}{20}$

$\frac{3}{10}$

P(Both fail) = P(John fail) and P(James fail)

And means multiplication (x)

P(Both fail) = P(John fail) x P(James fail)

Probability of not = 1 - Probability of is

P(John pass) = $\frac{3}{4}$

P(John fail) = 1 - P(John pass)

P(John fail) = 1 - $\frac{3}{4}$

Every number is being divided by 1.

1 = $\frac{1}{1}$

P(John fail) = $\frac{1}{1}$ - $\frac{3}{4}$

P(John fail) = $\frac{4 x 1 - 1 x 3}{4}$

P(John fail) = $\frac{4 - 3}{4}$

P(John fail) = $\frac{1}{4}$

P(James pass) = $\frac{3}{5}$

P(James fail) = 1 - P(James pass)

P(James fail) = 1 - $\frac{3}{5}$

P(James fail) = $\frac{1}{1}$ - $\frac{3}{5}$

P(James fail) = $\frac{5 x 1 - 1 x 3}{5}$

P(James fail) = $\frac{5 - 3}{5}$

P(James fail) = $\frac{2}{5}$

P(Both fail) = P(John fail) x P(James fail)

P(Both fail) = $\frac{1}{4}$ x $\frac{2}{5}$ = $\frac{1}{10}$

In a hall, there are 175 persons. 12% are children, 56 are men and the rest are women. If one person is selected at random from the hall, find the probability that a woman is selected.

$\frac{3}{25}$

$\frac{11}{25}$

$\frac{8}{25}$

$\frac{14}{25}$

Number from a percentage = Percentage x Total number

% means over 100

12% are children

Total number of persons = 175

Number of children = $\frac{12}{100}$ x 175

Number of children = 21

Number of women = Total number of persons - children - men

Number of women = 175 - 21 - 56

Number of women = 98

Probability = $\frac{Number of possible selection}{Number of samples}$

Probability (Woman) = $\frac{Number of women}{Number of persons}$

Probability (Woman) = $\frac{98}{175}$ = $\frac{14}{25}$

Note: 7 divides 98, 14 times and 175, 25 times.

A box contains 2 red, 6 white and 5 black balls, all of the same size. If a ball is selected at random, what is the probability that it is black?

$\frac{2}{13}$

$\frac{5}{11}$

$\frac{5}{13}$

$\frac{11}{13}$

Probability = $\frac{Number of possible selection}{Number of samples}$

P(Black) = $\frac{Number of black balls}{Total number of balls}$

Number of black balls = 5

Total number of balls = 2 + 6 + 5

Total number of balls = 13

P(Black) = $\frac{5}{13}$

A bag contains 32 blue, 28 black and 20 red identical balls of same size. If a ball is picked at random, what is the probability that it is not red?

$\frac{1}{4}$

$\frac{2}{5}$

$\frac{3}{5}$

$\frac{3}{4}$

Probability = $\frac{Number of possible selection}{Number of samples}$

Probability of is + Probability of not = 1

You can find the probability of picking a red ball and subtract it from 1 to get the not or find the total number of balls which are not red and divide by the total balls.

Total balls = 32 + 28 + 20
Total balls = 80

Method I

Number of red balls = 20

Probability of picking red ball = $\frac{20}{80}$

Probability of picking red ball = $\frac{1}{4}$

Probability of not red ball = 1 - $\frac{1}{4}$

Every number is being divided by 1.

1 = $\frac{1}{1}$

Probability of not red ball = $\frac{1}{1}$ - $\frac{1}{4}$

Probability of not red ball = $\frac{1x 4 - 1}{4}$

Probability of not red ball = $\frac{4 - 1}{4}$

Probability of not red ball = $\frac{3}{4}$

Method II

Balls which are not red = 32 + 28
Balls which are not red = 60

Probability of not red ball = $\frac{60}{80}$

Probability of not red ball = $\frac{3}{4}$

A die is rolled once. Find the probability of obtaining a number less than 3.

$\frac{1}{6}$

$\frac{5}{6}$

$\frac{2}{3}$

$\frac{1}{3}$

Probability = $\frac{Number of possible selection}{Number of samples}$

Numbers in a die rolled once = {1,2,3,4,5,6}

Numbers less than 3 = {1,2}

Number of possible selection = 2
Number of samples = 6

Probability of less than 3 = $\frac{2}{6}$ = $\frac{1}{3}$

The probabilities of three independent events X, Y and Z occurring are $\frac{1}{2}$ , $\frac{2}{3}$ and $\frac{1}{4}$ respectively. What is the probability that only events X and Z are occurring?

$\frac{1}{24}$

$\frac{3}{24}$

$\frac{1}{8}$

$\frac{3}{4}$

P(X) = $\frac{1}{2}$

P(Y) = $\frac{2}{3}$

P(Z) = $\frac{1}{4}$

P(only X and Z) = P(X) x P(Z) x P(Y not occurring)

Probability of not occurring = 1 - Probability of occurring

P(Y not occurring) = 1 - $\frac{2}{3}$

Every number is being divided by 1

P(Y not occurring) = $\frac{1}{1}$ - $\frac{2}{3}$

P(Y not occurring) = $\frac{3 x 1 - 1 x 2}{3}$

P(Y not occurring) = $\frac{3 - 2}{3}$

P(Y not occurring) = $\frac{1}{3}$

P(only X and Z) = $\frac{1}{2}$ x $\frac{1}{4}$ x $\frac{1}{3}$

P(only X and Z) = $\frac{1 x 1 x 1}{2 x 4 x 3}$

P(only X and Z) = $\frac{1}{24}$

A box contains 40 identical balls of which 10 are red and 12 are blue. If a ball is selected at random from the box, what is the probability that it is neither red nor blue?

$\frac{11}{20}$

$\frac{9}{20}$

$\frac{3}{10}$

$\frac{1}{4}$

Probability Formula

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of balls which are neither red nor blue = 40 - (10 + 12)

Number of balls which are neither red nor blue = 40 - 22

Number of balls which are neither red nor blue = 18

Number of samples = 40

P(Neither red nor blue) = $\frac{18}{40}$ = $\frac{9}{20}$

A fair die is tossed twice. What is the probability of getting a sum of at least 10?

$\frac{5}{36}$

$\frac{2}{3}$

$\frac{5}{18}$

$\frac{1}{6}$

1. Calculate the total number of outcomes when a fair die is tossed twice: 6 × 6 = 36

At least 10 means 10 or more.

2. Find the favorable outcomes for a sum of at least 10: (4,6), (6,4) for 10; (5,5) for 10; (5,6), (6,5) for 11, (6,6) for 12

Number of favourable outcome is 6

Probability Formula

Probability = $\frac{Number of possible selection}{Number of samples}$

P(Sum is at least 10) = $\frac{6}{36}$ = $\frac{1}{6}$

You can also draw a table for all the possible outcomes.

	1	2	3	4	5	6
1	1, 1	1, 2	1, 3	1, 4	1, 5	1, 6
2	2, 1	2, 2	2, 3	2, 4	2, 5	2, 6
3	3, 1	3, 2	3, 3	3, 4	3, 5	3, 6
4	4, 1	4, 2	4, 3	4, 4	4, 5	4, 6
5	5, 1	5, 2	5, 3	5, 4	5, 5	5, 6
6	6, 1	6, 2	6, 3	6, 4	6, 5	6, 6

Probability Formula

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 6 x 6 = 36

Number of possible selection = 6

P(Sum is at least 10) = $\frac{6}{36}$ = $\frac{1}{6}$

10.

A box contains 12 identical balls, of which 5 are red, 4 blue and the rest are green.

Use this information to answer the question below

If a ball is selected at random from the box, what is the probability that it is green?

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{3}{4}$

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 12

Number of green balls = 12 - (5 + 4)

Number of green balls = 12 - 9 = 3

P(Green) = $\frac{3}{12}$ = $\frac{1}{4}$

11.

A box contains 12 identical balls, of which 5 are red, 4 blue and the rest are green.

Use this information to answer the question below

If two balls are selected at random one after the other with replacement, what is the probability that both are red?

$\frac{103}{132}$

$\frac{5}{6}$

$\frac{5}{33}$

$\frac{25}{144}$

P(Both red) = P(first ball red) and P(second ball red)

And is multiplication (x)

P(Both red) = P(Red) x P(Red)

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 12

Number of red balls = 5

P(Red) = $\frac{5}{12}$

P(Both red) = $\frac{5}{12}$ x $\frac{5}{12}$ = $\frac{5 x 5}{12 x 12}$ = $\frac{25}{144}$

12.

There are 8 boys and 4 girls in a lift. What is the probability that the first person who steps out of the lift will be a boy?

$\frac{1}{4}$

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{3}{4}$

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 8 + 4 = 12

Number of possible selection = 8

P(Boy) = $\frac{8}{12}$ = $\frac{2}{3}$

13.

A box contains 2 white and 3 blue identical balls. If two balls are picked at random from the box, one after the other with replacement, what is the probability that they are of different colours.

$\frac{12}{25}$

$\frac{7}{20}$

$\frac{3}{5}$

$\frac{2}{3}$

Difference colours = First ball is white and the second ball is blue or the first ball is blue and the second ball is white.

P(Different colours) = P(White first) and P(Blue second) or P(Blue first) and P(White second)

And means multiplication (x) and or means addition (+)

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 2 + 3 = 5

Number of white balls = 2

P(White) = $\frac{2}{5}$

Number of blue balls = 3

P(Blue) = $\frac{3}{5}$

P(Different colours) = $\frac{2}{5}$ x $\frac{3}{5}$ + $\frac{3}{5}$ x $\frac{2}{5}$

P(Different colours) = $\frac{2 x 3}{5 x 5}$ + $\frac{3 x 2}{5 x 5}$

P(Different colours) = $\frac{6}{25}$ + $\frac{6}{25}$ = $\frac{6 + 6}{25}$ = $\frac{12}{25}$

14.

A box contains 5 red, 6 green and 7 yellow pencils of the same size. What is the probability of picking a green pencil at random?

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{1}{6}$

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 5 + 6 + 7 = 18

Number of possible selection for green pencil = 6

P(Green) = $\frac{6}{18}$ = $\frac{1}{3}$

15.

A fair die is thrown once. What is the probability of obtaining 3?

$\frac{1}{6}$

$\frac{1}{3}$

$\frac{1}{2}$

Samples = {1, 2, 3, 4, 5, 6}

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 6

Number of possible selection/Number of 3 = 1

P(3) = $\frac{1}{6}$

16.

In a certain community, 1 out of every 8 persons is a graduate. If Yoni and Etteh are members of the community, what is the probability that they are both graduates?

$\frac{1}{64}$

$\frac{1}{16}$

$\frac{1}{8}$

$\frac{1}{4}$

1 out of every 8 persons is a graduate

P(Graduate) = $\frac{1}{8}$

P(Yonni and Etteh Graduates) = P(Yonni being Graduate) and P(Etteh being Graduate)

And means multiplication (x)

P(Yonni and Etteh Graduates) = P(Yonni being Graduate) x P(Etteh being Graduate)

P(Yonni being Graduate) = $\frac{1}{8}$

P(Etteh being Graduate) = $\frac{1}{8}$

P(Yonni and Etteh Graduates) = $\frac{1}{8}$ x $\frac{1}{8}$

P(Yonni and Etteh Graduates) = $\frac{1 x 1}{8 x 8}$ = $\frac{1}{64}$

17.

Out of 20 girls in a class, 12 like Music and 15 like movies. If a girl is selected at random from the class, what is the probability that she liked both Music and Movies?

$\frac{3}{10}$

$\frac{7}{20}$

$\frac{3}{5}$

$\frac{3}{4}$

Let x = number of students who liked both Music and Movies.

Coverig Music,

12 + 15 - x = 20

27 - 20 = x

7 = x

x = 7

Number of students who liked both Music and Movies = 7

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of students = 20

P(Both Music and Movies) = $\frac{7}{20}$

18.

A box contains 12 red, 8 yellow and 10 green pebbles, all of the same size. If two pebbles are selected at random one after the other, with replacement, find the probability that the first is green and the second is not a red pebble.

$\frac{1}{9}$

$\frac{1}{5}$

$\frac{15}{20}$

$\frac{17}{20}$

Total pebbles = 12 + 8 + 10

Total pebbles = 30

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Probability (First pebble is green) = $\frac{Number of green pebbles}{Total number of pebbles}$

Probability (First pebble is green) = $\frac{10}{30}$

Probability (First pebble is green) = $\frac{1}{3}$

Probability (Second pebble is not red) = $\frac{Number of non-red pebbles}{Total number of pebbles}$

Number of non-red pebbles = 8 + 10

Number of non-red pebbles = 18

Probability (Second pebble is not red) = $\frac{18}{30}$

Probability (Second pebble is not red) = $\frac{3}{5}$

Since the selection is with replacement, the two events are independent.

Note: and in probability means multiplication

Probability = Probability (First pebble is green) x Probability (Second pebble is not red)

Probability = $\frac{1}{3}$ x $\frac{3}{5}$

Probability = $\frac{1 x 3}{3 x 5}$

Probability = $\frac{3}{15}$

Probability = $\frac{1}{5}$

19.

A number is chosen at random from the set {13, 14, ...., 30}. What is the probability that it is a prime number?

$\frac{5}{16}$

$\frac{1}{3}$

$\frac{5}{18}$

$\frac{3}{8}$

Set = {13, 14, ...., 30}

Set = {13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}

Number of samples = 18

Prime numbers are numbers with only two factors (1 and itself). Thus only 1 and the number itself can divide the number without a remainder.

Set of prime numbers from the set = {13, 17, 19, 23, 29}

Number of prime numbers = 5

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

P(Prime) = $\frac{5}{18}$

20.

The probability that John and James pass an examination are $\frac{3}{4}$ and $\frac{3}{5}$ respectively.

Find the probability that both will fail.

$\frac{1}{10}$

$\frac{11}{20}$

$\frac{9}{20}$

$\frac{3}{10}$

P(Both fail) = P(John fail) and P(James fail)

And means multiplication (x)

P(Both fail) = P(John fail) x P(James fail)

Probability of not = 1 - Probability of is

P(John pass) = $\frac{3}{4}$

P(John fail) = 1 - P(John pass)

P(John fail) = 1 - $\frac{3}{4}$

Every number is being divided by 1.

1 = $\frac{1}{1}$

P(John fail) = $\frac{1}{1}$ - $\frac{3}{4}$

P(John fail) = $\frac{4 x 1 - 1 x 3}{4}$

P(John fail) = $\frac{4 - 3}{4}$

P(John fail) = $\frac{1}{4}$

P(James pass) = $\frac{3}{5}$

P(James fail) = 1 - P(James pass)

P(James fail) = 1 - $\frac{3}{5}$

P(James fail) = $\frac{1}{1}$ - $\frac{3}{5}$

P(James fail) = $\frac{5 x 1 - 1 x 3}{5}$

P(James fail) = $\frac{5 - 3}{5}$

P(James fail) = $\frac{2}{5}$

P(Both fail) = P(John fail) x P(James fail)

P(Both fail) = $\frac{1}{4}$ x $\frac{2}{5}$ = $\frac{1}{10}$

21.

In a hall, there are 175 persons. 12% are children, 56 are men and the rest are women. If one person is selected at random from the hall, find the probability that a woman is selected.

$\frac{3}{25}$

$\frac{11}{25}$

$\frac{8}{25}$

$\frac{14}{25}$

Number from a percentage = Percentage x Total number

% means over 100

12% are children

Total number of persons = 175

Number of children = $\frac{12}{100}$ x 175

Number of children = 21

Number of women = Total number of persons - children - men

Number of women = 175 - 21 - 56

Number of women = 98

Probability = $\frac{Number of possible selection}{Number of samples}$

Probability (Woman) = $\frac{Number of women}{Number of persons}$

Probability (Woman) = $\frac{98}{175}$ = $\frac{14}{25}$

Note: 7 divides 98, 14 times and 175, 25 times.

22.

A box contains 2 red, 6 white and 5 black balls, all of the same size. If a ball is selected at random, what is the probability that it is black?

$\frac{2}{13}$

$\frac{5}{11}$

$\frac{5}{13}$

$\frac{11}{13}$

Probability = $\frac{Number of possible selection}{Number of samples}$

P(Black) = $\frac{Number of black balls}{Total number of balls}$

Number of black balls = 5

Total number of balls = 2 + 6 + 5

Total number of balls = 13

P(Black) = $\frac{5}{13}$

23.

A bag contains 32 blue, 28 black and 20 red identical balls of same size. If a ball is picked at random, what is the probability that it is not red?

$\frac{1}{4}$

$\frac{2}{5}$

$\frac{3}{5}$

$\frac{3}{4}$

Probability = $\frac{Number of possible selection}{Number of samples}$

Probability of is + Probability of not = 1

You can find the probability of picking a red ball and subtract it from 1 to get the not or find the total number of balls which are not red and divide by the total balls.

Total balls = 32 + 28 + 20
Total balls = 80

Method I

Number of red balls = 20

Probability of picking red ball = $\frac{20}{80}$

Probability of picking red ball = $\frac{1}{4}$

Probability of not red ball = 1 - $\frac{1}{4}$

Every number is being divided by 1.

1 = $\frac{1}{1}$

Probability of not red ball = $\frac{1}{1}$ - $\frac{1}{4}$

Probability of not red ball = $\frac{1x 4 - 1}{4}$

Probability of not red ball = $\frac{4 - 1}{4}$

Probability of not red ball = $\frac{3}{4}$

Method II

Balls which are not red = 32 + 28
Balls which are not red = 60

Probability of not red ball = $\frac{60}{80}$

Probability of not red ball = $\frac{3}{4}$

24.

A die is rolled once. Find the probability of obtaining a number less than 3.

$\frac{1}{6}$

$\frac{5}{6}$

$\frac{2}{3}$

$\frac{1}{3}$

Probability = $\frac{Number of possible selection}{Number of samples}$

Numbers in a die rolled once = {1,2,3,4,5,6}

Numbers less than 3 = {1,2}

Number of possible selection = 2
Number of samples = 6

Probability of less than 3 = $\frac{2}{6}$ = $\frac{1}{3}$

25.

$\frac{1}{24}$

$\frac{3}{24}$

$\frac{1}{8}$

$\frac{3}{4}$

P(X) = $\frac{1}{2}$

P(Y) = $\frac{2}{3}$

P(Z) = $\frac{1}{4}$

P(only X and Z) = P(X) x P(Z) x P(Y not occurring)

Probability of not occurring = 1 - Probability of occurring

P(Y not occurring) = 1 - $\frac{2}{3}$

Every number is being divided by 1

P(Y not occurring) = $\frac{1}{1}$ - $\frac{2}{3}$

P(Y not occurring) = $\frac{3 x 1 - 1 x 2}{3}$

P(Y not occurring) = $\frac{3 - 2}{3}$

P(Y not occurring) = $\frac{1}{3}$

P(only X and Z) = $\frac{1}{2}$ x $\frac{1}{4}$ x $\frac{1}{3}$

P(only X and Z) = $\frac{1 x 1 x 1}{2 x 4 x 3}$

P(only X and Z) = $\frac{1}{24}$

26.

A box contains 40 identical balls of which 10 are red and 12 are blue. If a ball is selected at random from the box, what is the probability that it is neither red nor blue?

$\frac{11}{20}$

$\frac{9}{20}$

$\frac{3}{10}$

$\frac{1}{4}$

Probability Formula

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of balls which are neither red nor blue = 40 - (10 + 12)

Number of balls which are neither red nor blue = 40 - 22

Number of balls which are neither red nor blue = 18

Number of samples = 40

P(Neither red nor blue) = $\frac{18}{40}$ = $\frac{9}{20}$

27.

A box contains 12 identical balls, of which 5 are red, 4 blue and the rest are green.

Use this information to answer the question below

If a ball is selected at random from the box, what is the probability that it is green?

$\frac{1}{4}$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{3}{4}$

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 12

Number of green balls = 12 - (5 + 4)

Number of green balls = 12 - 9 = 3

P(Green) = $\frac{3}{12}$ = $\frac{1}{4}$

28.

A box contains 12 identical balls, of which 5 are red, 4 blue and the rest are green.

Use this information to answer the question below

If two balls are selected at random one after the other with replacement, what is the probability that both are red?

$\frac{103}{132}$

$\frac{5}{6}$

$\frac{5}{33}$

$\frac{25}{144}$

P(Both red) = P(first ball red) and P(second ball red)

And is multiplication (x)

P(Both red) = P(Red) x P(Red)

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 12

Number of red balls = 5

P(Red) = $\frac{5}{12}$

P(Both red) = $\frac{5}{12}$ x $\frac{5}{12}$ = $\frac{5 x 5}{12 x 12}$ = $\frac{25}{144}$

29.

There are 8 boys and 4 girls in a lift. What is the probability that the first person who steps out of the lift will be a boy?

$\frac{1}{4}$

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{3}{4}$

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 8 + 4 = 12

Number of possible selection = 8

P(Boy) = $\frac{8}{12}$ = $\frac{2}{3}$

30.

A box contains 2 white and 3 blue identical balls. If two balls are picked at random from the box, one after the other with replacement, what is the probability that they are of different colours.

$\frac{12}{25}$

$\frac{7}{20}$

$\frac{3}{5}$

$\frac{2}{3}$

Difference colours = First ball is white and the second ball is blue or the first ball is blue and the second ball is white.

P(Different colours) = P(White first) and P(Blue second) or P(Blue first) and P(White second)

And means multiplication (x) and or means addition (+)

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 2 + 3 = 5

Number of white balls = 2

P(White) = $\frac{2}{5}$

Number of blue balls = 3

P(Blue) = $\frac{3}{5}$

P(Different colours) = $\frac{2}{5}$ x $\frac{3}{5}$ + $\frac{3}{5}$ x $\frac{2}{5}$

P(Different colours) = $\frac{2 x 3}{5 x 5}$ + $\frac{3 x 2}{5 x 5}$

P(Different colours) = $\frac{6}{25}$ + $\frac{6}{25}$ = $\frac{6 + 6}{25}$ = $\frac{12}{25}$

31.

A box contains 5 red, 6 green and 7 yellow pencils of the same size. What is the probability of picking a green pencil at random?

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{1}{6}$

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 5 + 6 + 7 = 18

Number of possible selection for green pencil = 6

P(Green) = $\frac{6}{18}$ = $\frac{1}{3}$

32.

A fair die is thrown once. What is the probability of obtaining 3?

$\frac{1}{6}$

$\frac{1}{3}$

$\frac{1}{2}$

Samples = {1, 2, 3, 4, 5, 6}

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of samples = 6

Number of possible selection/Number of 3 = 1

P(3) = $\frac{1}{6}$

33.

In a certain community, 1 out of every 8 persons is a graduate. If Yoni and Etteh are members of the community, what is the probability that they are both graduates?

$\frac{1}{64}$

$\frac{1}{16}$

$\frac{1}{8}$

$\frac{1}{4}$

1 out of every 8 persons is a graduate

P(Graduate) = $\frac{1}{8}$

P(Yonni and Etteh Graduates) = P(Yonni being Graduate) and P(Etteh being Graduate)

And means multiplication (x)

P(Yonni and Etteh Graduates) = P(Yonni being Graduate) x P(Etteh being Graduate)

P(Yonni being Graduate) = $\frac{1}{8}$

P(Etteh being Graduate) = $\frac{1}{8}$

P(Yonni and Etteh Graduates) = $\frac{1}{8}$ x $\frac{1}{8}$

P(Yonni and Etteh Graduates) = $\frac{1 x 1}{8 x 8}$ = $\frac{1}{64}$

34.

Out of 20 girls in a class, 12 like Music and 15 like movies. If a girl is selected at random from the class, what is the probability that she liked both Music and Movies?

$\frac{3}{10}$

$\frac{7}{20}$

$\frac{3}{5}$

$\frac{3}{4}$

Let x = number of students who liked both Music and Movies.

Coverig Music,

12 + 15 - x = 20

27 - 20 = x

7 = x

x = 7

Number of students who liked both Music and Movies = 7

Probability

Probability = $\frac{Number of possible selection}{Number of samples}$

Number of students = 20

P(Both Music and Movies) = $\frac{7}{20}$

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