In this article, we will discuss how to solve problems related to probability. But before proceeding to solve the problems, first, let us see what is a probability.

In mathematics, probability refers to the possibility of the occurrence of a random event. The value of probability ranges from 0 to 1 and it is never expressed as a negative number. It is a branch of mathematics that deals with the forecast of the events. To calculate the probability of a random event, you must know the total number of outcomes.

Why we use Probability?

We cannot predict many events with total certainty. For example, we cannot say that it will certainly rain tomorrow. Instead, we can say that there is a 90% chance that it will rain tomorrow. As you have seen that we have predicted the rain in percentage. This is because we can convert the probability into a percentage by multiplying it by 100%. In terms of probability, we can say that the probability that it will rain tomorrow is 0.9. If there is a 100% chance that an event will occur, then the probability of such an event is 1. The probability of an impossible event is 0.

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Probability Formula

The formula to calculate the probability of an event is given below:

Probability = \frac {Number \hspace{0.5cm} of \hspace{0.5cm}favorable \hspace{0.5cm} outcomes} {Total \hspace{0.5cm} number\hspace{0.5cm}of \hspace{0.5cm}outcomes}

 

In the next section, we will solve some examples related to probability.

Example 1

The probability of John attending an event is 0.7. What is the probability that he will not attend the event?

Solution

The probability of attending an event = 0.7

The probability of not attending an event = 1 - 0.7 = 0.3

 

Example 2

There are 12 balls in a bag. Out of 12 balls, 3 are red, 4 are blue, and 5 are green. What is the probability of picking up:

a) a red ball

b) a red or green ball

c) a red and blue ball

Solution

There are three parts of this problem. We will solve each part step by step.

Part a

Number of red balls in a bag = 3

Total number of balls in a bag = 12

Probability of picking up a red ball = \frac {3}{12} = \frac {1}{4}

Part b

Number of red balls in a bag = 3

Number of green balls in a bag = 5

Total number of balls in a bag = 12

Probability of picking up a red ball = P(R) = \frac {3}{12} = \frac {1}{4}

Probability of picking up a green ball = P(G) = \frac {5}{12}

Probability of picking up a red or green ball = P(R) U P(G)

= \frac{1}{4} + \frac{5}{12}

= \frac {2}{3}

 

Part c

Probability of picking up a red ball = P(R) = \frac {3}{12} = \frac {1}{4}

Probability of picking up a blue ball = P(B) = \frac {4}{12} = \frac {1}{3}

Probability of picking up a red and blue ball = 0

We cannot pick up both the balls at the same time. So, both the events are mutually exclusive. The probability of mutually exclusive events is always zero.

 

Example 3

From a deck of 52 cards, what is the probability of picking up a queen or jack card?

Solution

Total number of cards in a deck = 52

Number of queen cards=  4

Probability of picking up a queen card = P(Q) = \frac {4}{52} = \frac{1}{13}

Number of jack cards = 4

The probability of picking up a jack card = P(J) =  \frac {4}{52} = \frac {1}{13}

The probability of picking up a queen or a jack card = P(Q) U P(J)

=\frac{1}{13} + \frac{1}{13}

=\frac{2}{13}

 

Example 4

A box contains 5 red, 8 green, 12 blue, and 10 yellow marbles. If three marbles are drawn from the box at random, what is the probability that the first marble will be red, second will be yellow, and third will be green?

Solution

Total number of marbles in a box = 5 + 8 + 12 + 10 = 35

Number of red marbles = 5

Probability of picking up a red marble first = P(R) = \frac {5}{35} = \frac {1}{7}

Number of marbles left = 35 - 1 = 34

Number of yellow marbles in a box = 10

Probability of picking up a second yellow marble = P(Y) = \frac {10}{34} = \frac {5}{17}

Number of marbles left = 34 - 1 = 33

Number of green marbles = 8

Probability of picking up a third green marble = P(G) = \frac {8}{33}

 

Example 5

Two dice are rolled simultaneously. What is the probability that the sum is:

a) equal to 5

b) less than 5

c) greater than 5

Solution

Part a

Total number of outcomes = 6 x 6 = 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) }

Let A be the event "sum equal to 5". Number of possible outcomes with sum equal to 5 are:
(1,4), (2,3), (3,2), (4,1)
Number of favorable outcomes = 4
Total number of outcomes = 36
Probability that the sum of two number on the dice is equal to 5 = P(A) = \frac {4}{36} = \frac{1}{9}

Part b

Let B be the event "sum less than 5". Number of possible outcomes with sum less than 5 are:
(1,1), (1,2), (1,3), (2,1), (2,2), (3,1)
Number of favorable outcomes = 6
Total number of outcomes = 36
Probability that the sum of two number on the dice is greater than 5 = P(B) = \frac {6}{36} = \frac{1}{6}

Part c

Let C be the event "sum greater than 5". Number of possible outcomes with sum greater than 5 are:
(1,5), (1,6), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6), (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 favorable outcomes = 26
Total number of outcomes = 36
Probability that the sum of two number on the dice is greater than 5 = P(C) = \frac {26}{36} = \frac{13}{18}

Example 6

From a deck of 52 cards, 2 cards are chosen at random. What is the probability of:
a) picking up a black card first
b) picking up a second red card

Solution

Part a

Total number of cards = 52
Number of black cards = 26
Picking up a first black card = \frac {26}{52} = \frac {1}{2}

Part b

Number of cards left = 52 - 1 = 51
Picking up a second red card = \frac {26}{51}

Example 7

A coin is tossed 4 times. What is the probability of getting 4 consecutive heads?

Solution

Probability of getting head single time = \frac {1}{2}
Probability of getting head four times = \frac {1}{2} \times \frac {1}{2} \times \frac {1}{2} \times \frac {1}{2}
                                                                      = \frac {1}{16}

Example 8

Three blocks are chosen at random with replacement from a box that contains 5 black, 6 white and 9 blue blocks. What is the probability of picking up a black block all the three times?

Solution

It is given that the blocks are chosen and then replaced again.
Total number of outcomes = 5 + 6 + 9 = 20
Number of black blocks = 5
Probability of picking up a black block one time = \frac {5}{20} = \frac {1}{4}
Probability of picking up a black block three times = \frac {1}{4} \times \frac {1}{4} \times \frac {1}{4}
                                                                                             =\frac {1}{64}

Example 9

From a deck of 52 cards, what is the probability of picking up:
a) a face card first time
b) a face card second time

Solution

Total number of cards in a deck = 52
Number of face cards = 4 queen + 4 king + 4 jack = 12 cards
Probability of picking up a face card first time = \frac {12}{52} = \frac {3}{13}
Total number of cards left = 52 - 1 = 51
Number of face cards left = 12 - 1 = 11
Probability of picking up a face card second time = \frac {11}{51}

Example 10

A dice and coin is thrown simultaneously. What is the probability of getting a number greater than 4 on a dice and tail on the coin?

Solution

Total number of outcomes for a dice = 6
Number of favorable outcomes (getting a number greater than 4) = 2
Probability of getting a number greater than 4 = \frac{2}{6} = \frac {1}{3}
Total number of outcomes for a coin = 2
Number of favorable outcomes (getting a tail) = 1
Probability of getting a tail = \frac {1}{2}
Probability of getting a number greater than 4 on a dice and tail on a coin = \frac {1}{3} \times \frac {1}{2}
                                                                                                                                       = \frac {1}{6}
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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.