Exercise 1

Two balls are drawn from a box that contains three blue, four red, one green and one yellow ball. What is the probability of selecting:

1  The first red ball

2 The second blue ball without replacement

 

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Exercise 2

A box has 8 red balls, 5 yellow and 7 green. If a ball is extracted at random, calculate the probability that it will be:

1 Red

2 Green

3 Yellow

4 Not red

5 Not yellow

 

Exercise 3

Two dice are thrown into the air simultaneously. Calculate:

  • The probability that the sum of the dice will total 7.
  • The probability that the sum of the dice will be an even number.
  • The probability that the sum of the dice will be a multiple of three.

 

Exercise 4

Three dice are thrown simultaneously. Calculate the probability of:

  • Obtaining a six in all three die.
  • Obtaining the same numbers on all the three dice

 

Exercise 5

Calculate the probability of obtaining the following outcomes when rolling a die:

  • An even number
  • A multiple of three
  • A five or a six

 

Exercise 6

A worker at the factory accidentally mixes seven good and two defective lamps. To find the defective lamps, he tests each lamp one by one, randomly without replacement. What is the probability that he is lucky and find both the defective lamps in the first two attempts?

Solution of Exercise 1

Total number of balls in a bag = 9

Probability of selecting the first red ball = P(A) = \frac{4}{9}

Probability of selecting the second blue ball = P(B) = \frac{3}{8}

Solution of Exercise 2

Probability of selecting a red ball = P(A) = \frac{8}{20} = \frac{2}{5}

Probability of selecting a green ball = P(B) = \frac{7}{20}

Probability of selecting a yellow ball = P(C) = \frac{5}{20} = \frac{1}{4}

Probability of selecting a ball that is not red = P(D) = \frac{12}{20} = \frac{3}{5}

Probability of selecting a ball that is not yellow = P(E) = \frac{15}{20} = \frac{3}{4}

 

Solution of Exercise 3

Total number of outcomes of two dice = 36

Sample space = S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), (4,1),(4,2), (4,3),(4,4),(4,5),(4,6), (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Number of events in which the total sum of the numbers on the dice will be 7 = (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

Probability of getting the sum 7 on two dice = \frac{6}{36} = \frac{1}{6}

Number of events in which the sum of the numbers on the dice is multiple of 3 = (1,2), (1,5), (2,1), (2,4), (3,3), (3,6), (4,2), (4,5), (5,1), (5,4), (6,3), (6,6)

Probability of getting a sum that is multiple of 3 = \frac{12}{36} = \frac{1}{3}

Solution of Exercise 4

Total number of outcomes of three dice = 6 x 6 x 6 = 216

Probability of obtaining 6 on all the three dice = \frac{1}{216}

Probability of obtaining the same numbers on all the three dice = \frac{6}{216} = \frac{1}{36}

 

Solution of Exercise 5

Probability of getting an even number on a die = \frac{3}{6} = \frac{1}{2}

Probability of getting a multiple of 3 on a die = \frac{2}{6} = \frac{1}{3}

Probability of getting a five or a six = \frac{2}{6} = \frac{1}{3}

 

Solution of Exercise 6

Suppose A_1 and A_2 are the events that he finds the defective lamps.

The probability of selecting the first defective lamp = \frac{2}{9}

The probability that the second lamp is also defective given the first one is defective = P(A_1 \cap A_2) = \frac{1}{8}

We know that we have to find the probability that both the lamps are defective. We will use the following formula of conditional probability:

P(A_1|A_2) = \frac{P(A_1 \cap A_2)}{P(A_1)}

We have to find only P(A_1 \cap A_2) because we already have P(A_1 \cap A_2).

P(A_1 \cap A_2) = P(A_1|A_2) \cdot P(A_1)

= \frac{1}{8} \cdot \frac{1}{9}

= \frac{1}{72}

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.