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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: The probability of a certain event is 1 and an impossible event is 0. The probability is always positive. The best Maths tutors available
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1st lesson free!  5 (17 reviews)
Matthew
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1st lesson free!  4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!  4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!  5 (28 reviews)
Ayush
£60
/h
1st lesson free!  4.9 (9 reviews)
Petar
£27
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1st lesson free!  4.9 (11 reviews)
Rajan
£15
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1st lesson free!  5 (13 reviews)
Farooq
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## 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

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

= = ### Part c

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

### Part a

Total number of cards = 52
Number of even-numbered cards = 4 x 5 = 20
Picking up a first even-numbered card = ### Part b

Number of cards left = 52 - 1 = 51
Picking up a second even-numbered card = ### Part c

Number of cards left = 51 - 1 = 50
Picking up a third even-numbered card = ## Example 3

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

### Solution

Probability of getting tail single time = Probability of getting tail five times = = ## 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 = Total number of outcomes for a coin = 2
Number of favorable outcomes (getting a head) = 1
Probability of getting a head P(H) = Probability of getting a prime number on a dice and head on a coin = = ## 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 ## 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

### 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) = ### 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) = ### 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) = ## 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 = 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 = 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: = ## 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

### Part a

Total number of cards = 52
Number of ace cards = 4
Picking up a first ace card = ### 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 = ### 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 = Need a Maths teacher?

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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.