In this article, we will see how to solve different types of probability word problems. But before proceeding to examples, let us define the probability first.

What is Probability?

Probability can be defined as:

The possibility of the occurrence of a random event

The formula for calculating the probability 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}

 

The probability of a certain event is 1 and an impossible event is 0. The probability is always positive.

 

The best Maths tutors available
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Ayush
5
5 (28 reviews)
Ayush
£60
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Ayush
5
5 (28 reviews)
Ayush
£60
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
First Lesson Free>

Example 1

There are 50 cookies in a box. Out of 50 cookies, 15 are chocolate-flavored, 20 are caramel-flavored, and the remaining are salted cookies. What is the probability of picking up:

a) a chocolate cookie randomly

b) a chocolate or caramel-flavored cookie

c) a chocolate and salted cookie

Solution

Part a

Number of chocolate-flavored cookies in a box = 15

Total number of cookies in a box = 50

Probability of picking up chocolate-flavored cookie = \frac {15}{50} = \frac {3}{10}

Part b

Number of chocolate-flavored cookies in a box = 15

Number of caramel-flavored cookies in a box = 20

Total number of cookies in a box = 50

Probability of picking up chocolate-flavored cookie = P(A) = \frac {15}{50} = \frac {3}{10}

Probability of picking up a caramel-flavored cookie = P(B) = \frac {20}{50} = \frac {2}{5}

Probability of picking up a chocolate or caramel cookie = P(A) U P(B)

= \frac{3}{10} + \frac{2}{5}

= \frac {7}{10}

 

Part c

Probability of picking up a chocolate cookie = P(A) = \frac {3}{10}

Probability of picking up a caramel= P(B) = \frac {2}{5}

Probability of picking up a chocolate or caramel cookie= P(R) \cap P(B)

Both events are mutually exclusive, so the probability of picking up a chocolate and caramel cookie is 0.

 

Example 2

From a deck of 52 cards, 3 cards are chosen at random. What is the probability of:
a) picking up an even-numbered card first
b) picking up a second even-numbered card
c) picking up a third even-numbered card

Solution

Part a

Total number of cards = 52
Number of even-numbered cards = 4 x 5 = 20
Picking up a first even-numbered card = \frac {20}{52} = \frac {5}{13}

Part b

Number of cards left = 52 - 1 = 51
Picking up a second even-numbered card = \frac {19}{51}

Part c

Number of cards left = 51 - 1 = 50
Picking up a third even-numbered card = \frac {18}{50} = \frac {9}{25}

Example 3

A coin is tossed 5 times. What is the probability of getting 5 consecutive tails?

Solution

Probability of getting tail single time = \frac {1}{2}
Probability of getting tail five times = \frac {1}{2} \times \frac {1}{2} \times \frac {1}{2} \times \frac {1}{2} \cdot \frac {1}{2}
                                                                      = \frac {1}{32}

Example 4

A dice and coin are thrown simultaneously. What is the probability of getting a prime number on a dice and head on the coin?

Solution

Total number of outcomes for a dice = 6
Number of favorable outcomes (getting a prime number) = 3
Probability of getting a prime number = \frac{3}{6} = \frac {1}{2}
Total number of outcomes for a coin = 2
Number of favorable outcomes (getting a head) = 1
Probability of getting a head P(H) = \frac {1}{2}
Probability of getting a prime number on a dice and head on a coin = \frac {1}{2} \times \frac {1}{2}
                                                                                                                                       = \frac {1}{4}

Example 5

In an area, 47% of the families own a car and a bike. 53% of the families own a car. What is the probability of a family owning a bike given that it already owns a car?

Solution

Percentage of the families who own both car and a bike = 47%

Probability of families owning a car and a bike= P(C and B) = 0.47

Percentage of families owning a car = 53%

The probability of families owning a car only = P(C) = 0.53

P(B|C) = \frac {P (C\hspace{0.3cm} and\hspace{0.3cm} B)} {P(C)} = \frac{0.47}{0.53} = 0.88

 

 

Example 5

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

a) less than 10

b) equal to 4

c) greater than 10

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 less than 10". The number of possible outcomes with the sum less than 10 are:
(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), (5,1),(5,2),(5,3),(5,4), (6,1),(6,2),(6,3)
Number of favorable outcomes = 30
Total number of outcomes = 36
Probability that the sum of two number on the dice is equal to 5 = P(A) = \frac {30}{36} = \frac{5}{6}

Part b

Let B be the event "sum equal to 4". Number of possible outcomes with sum equal to 4 are:
(1,3), (2,2), (3,1)
Number of favorable outcomes = 3
Total number of outcomes = 36
Probability that the sum of two number on the dice is equal to 4 = P(B) = \frac {3}{36} = \frac{1}{12}

Part c

Let C be the event "sum greater than 10". The number of possible outcomes with the sum greater than 10 are:
(5, 6), (6, 5), (6,6)
Number of favorable outcomes = 3
Total number of outcomes = 36
Probability that the sum of two number on the dice is greater than 10 = P(C) = \frac {3}{36} = \frac{1}{12}

Example 6

What is the probability of selecting two prime number cards from the deck of 52 cards?

Solution

First, we will evaluate the types of cards present in each deck. A deck of 52 cards has the following attributes:

  • There are four suits in a deck. Each suit has 13 cards
  • Two of the four suits are black cards and two are red.
  • The 13 cards in each suit include a king, a queen, a jack, ace, 2, 3, 4, 5, 6, 7, 8, 9, and 10.

In each suit, there are four prime numbers 2, 3, 5, and 7. Hence, in four suits, there will be 16 cards

Number of prime number cards in a deck = 16

Total number of cards in a deck = 52

The probability of selecting a prime number card from a deck = \frac{16}{52} = \frac{4}{13}

 

Now, we will calculate the probability of picking up the second prime number card from the deck.

Number of prime number cards left in a deck = 15

Total number of cards left in a deck = 51

The probability of drawing a prime number card, given a prime number card has already been drawn = \frac{15}{51}

 

In this problem, we need to compute the probability of picking up two prime number cards, therefore we will multiply the probabilities we have calculated above:

P(A \hspace{0.3cm} and \hspace{0.3cm} B) = P(A) \cdot P(B|A) = \frac{4}{13} \cdot \frac{15}{51}

= \frac{60} {663} = \frac{20}{221}

 

Example 7

From a deck of 52 cards, 3 cards are chosen at random. What is the probability of:
a) picking up an ace card first
b) picking up a second even-numbered card
c) picking up a third prime number card

Solution

Part a

Total number of cards = 52
Number of ace cards = 4
Picking up a first ace card = \frac {4}{52} = \frac {1}{13}

Part b

Number of cards left = 52 - 1 = 51
Number of even-numbered cards in a deck = 20
Picking up a second even-numbered card = \frac {20}{51}

Part c

Number of cards left = 51 - 1 = 50
Number of prime number cards in a deck = 16
Probability of picking up a third prime number card = \frac {16}{50} = \frac {8}{25}

 

 

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 5.00/5 - 1 vote(s)
Loading...

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.