Probability exercises solved (easy)

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The probability of a given result occurring in a random experiment is expressed through the ratio:

straight P space equal to space numerator straight no space space possibilities favorable space over straight denominator no space total space space space possibilities end of fraction

Next we have 10 questionseasy level solved About the subject. After the template we prepare comments that will show you how to perform the calculations.

question 1

If we roll a die, what is the probability of getting a number greater than 4?

a) 2/3
b) 1/4
c) 1/3
d) 3/2

Correct answer: c) 1/3

A die has 6 sides with numbers from 1 to 6. Therefore, the number of possibilities at launch is 6.

An event favorable to choosing a number greater than 4 is getting 5 or 6, that is, there are two possibilities.

Therefore, the probability that a number greater than 4 is the result of rolling the die is given for the reason:

straight P space equal to space 2 over 6 space equal to space 1 third

question 2

If we flip a coin, what is the probability of the “heads” side facing up?

a) 1/3
b) 1/2
c) 1/4
d) 0

Correct answer: b) 1/2

When tossing a coin, there are only two possibilities: flipping a coin. If the event of interest is "head", then the probability of it occurring is given by:

straight P space equals space 1 half space equals space 50 percent sign
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question 3

A restaurant has 13 people: 9 customers and 4 waiters. If we randomly pick a local person, what is the probability of being a customer?

a) 3/13
b) 9/13
c) 6/13
d) 7/13

Correct answer: b) 9/13.

If the favorable event is getting a customer, then the number of possibilities is 9.

As the restaurant has a total of 13 people, the probability of randomly choosing a customer is given by:

straight P space equal to space 9 over 13

question 4

If you randomly choose a letter in the alphabet, what is the probability of selecting a vowel?

a) 5/13
b) 7/13
c) 7/26
d) 5/26

Correct answer: d) 5/26

The alphabet has 26 letters, of which 5 are vowels. So the probability is:

straight P space equal to space 5 over 26

question 5

If a number from the sequence (2, 3, 5, 7, 11, 13, 17, 19) is randomly chosen, what is the probability of choosing a prime number?

a) 3/8
b) 1
c) 0
d) 5/8

Correct answer: b) 1

All 8 numbers in the sequence are prime numbers, that is, they are divisible only by the number 1 and by itself. Therefore, the probability of choosing a prime number in the sequence is:

straight P space equal to space 8 over 8 equal to space 1

question 6

If a class consists of 8 female and 7 male students and the teacher chooses randomly a student to go to the board to solve an exercise, what is the probability of being selected a student?

a) 8/15
b) 7/15
c) 11/15
d) 13/15

Correct answer: a) 8/15

The total number of students in the class is 15, 8 females and 7 males. As the favorable event is choosing a student, there are 8 possibilities of choice and the probability is given by:

straight P space equal to numerator space 8 over denominator 15 end of fraction

question 7

By randomly choosing a day of the week, what is the probability of choosing a Monday or a Friday?

a) 4/7
b) 1/7
c) 2/7
d) 3/7

Correct answer: c) 2/7.

The week is made up of 7 days.

The probability of choosing a Monday is 1/7 and the probability of choosing a Friday is also 1/7.

Therefore, the probability of choosing Monday or Friday is:

straight P space equal to space 1 over 7 space plus space 1 over 7 space equal to space 2 over 7

question 8

One person went to the bakery to buy bread and yogurt. If the establishment has 30 breads, 5 of which are from the previous day and the others were made on the day, and 20 yoghurts with the date of ineligible validity, of which 1 has already expired, what is the probability of the customer choosing a bread of the day and a yogurt within the validity?

a) 19/24
b) 17/30
c) 14/27
d) 18/29

Correct answer: a) 19/24

If the bakery has 30 loaves and 25 are not from the day before, then the probability of choosing a loaf of the day is given by:

straight P with 1 subscript space equal to space 25 over 30 space equal to space 5 over 6

If there is an expired yogurt among the 20 units of the bakery, then the probability of choosing a yogurt within the expiration date is:

straight P with 2 subscript space equal to 19 over 20

Therefore, the probability of choosing a bread of the day and a yogurt within the validity period is:

straight P with 1 subscript straight space x straight space P with 2 subscript space equal to space 5 over 6 straight space x space 19 over 20 space equal to space numerator 5 straight x 19 over denominator 6 straight x 20 end of fraction equal to space 95 over 120 space equal to 19 about 24

question 9

João has a jar with colored candies. One day he decided to count how many candies of each color were in the container and came up with the numbers:

  • 6 red bullets
  • 3 green bullets
  • 5 white bullets
  • 7 yellow bullets

Putting all the candies back in the jar and choosing two candies to eat, what is the probability that John will randomly pick up a red and a yellow candy?

a) 4/19
b) 3/27
c) 1/23
d) 2/21

Answer: d) 2/21

The total number of bullets in the pot is: 6+3+5+7 = 21

The probability of catching a red bullet is given by:

straight P space equal to space 6 over 21

The probability of choosing a yellow candy is:

straight P space equal to space 7 over 21

Therefore, the probability of choosing a red and a yellow candy is:

straight P space equal to space 6 over 21 space space x space 7 over 21 space equal to numerator space 6 x 7 over denominator 21 x 21 end of fraction space equal to space numerator 42 over denominator 441 space end of fraction equal to 2 about 21

question 10

What is the probability of choosing a card from the deck and that card is not an ace?

a) 12/17
b) 12/13
c) 14/13
d) 12/11

Answer: b) 12/13

A deck consists of 52 cards, of which 4 are ace, one in each suit.

So the probability of picking an ace is straight P space equal to 4 over 52.

The probability of not picking an ace is:

straight P ’ space equal to space 1 space – straight space P space straight space P ’ space equal to space 1 space – space numerator 4 over denominator 52 space end of fraction straight space P ’ space equal to space numerator 52 space – space 4 over denominator 52 end of fraction straight space P ’ space equal to space 48 over 52 equal to space 12 about 13

Get more knowledge with the contents:

  • Concept and calculation of probability
  • conditional probability
  • Probability Exercises
  • combinatorial analysis
  • Exercises on combinatorial analysis
  • Permutation
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