In this article, we will solve some examples related to the probability. So, let us get started.

Example 1

Different types of fasteners are present in a tool kit. There are 75 nuts, 50 screws and 20 bolts in it. Find the probability by considering the given scenarios:

a) A randomly selected fastener is a nut or a bolt.

b) Choosing two nuts from the tool kit without replacement

c) Choosing a screw first, and then a bolt when the first screw was replaced back

d) Choosing two bolts without replacement

Solution

We will solve the parts of this example one by one.

Part a

Total number of fasteners present in the tool kit = 75 + 50 + 20 = 145

Number of nuts in the tool kit = 75

Number of bolts in a tool kit = 20

Probability that a randomly selected fastener is a nut or a bolt = \frac{75 + 20} {145} = \frac{95}{145} = \frac{19}{29}

 

Part b

Number of nuts in a tool kit = 75

Total number of fasteners in the tool kit = 145

Probability of choosing first nut = P(A) = \frac{75}{145}

Probability of choosing the second nut from the tool kit = \frac{74}{144}

Probability of choosing the two nuts from the tool kit without replacement = \frac{75}{145} \times \frac{74}{144} =\frac{185}{696}

 

Part c

Number of screws in the tool kit = 50

Total number of fasteners in the tool kit = 145

Probability of choosing the first screw = \frac{50}{145}

Probability of choosing the second bolt with replacement = \frac{20}{145}

The probability of choosing a screw first and then a bolt = \frac{50}{145} \cdot \frac{20}{145} = \frac{40}{841}

 

Part d

Probability of choosing the first bolt = \frac{20}{145}

Probability of choosing the second bolt = \frac{19}{144}

The probability of choosing two bolts without replacement = \frac{20}{145} \cdot \frac{19}{144} = \frac{19}{1044}

 

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Example 2

John wants to go to a restaurant on a weekend. The probabilities of going to a restaurant on a certain day with or without rain are 0.64 and 0.25 respectively. If the probability that it is going to rain on a weekend is 0.45, then determine the probability that John will go to the restaurant?

Solution

Suppose A be the event that John will go the restaurant and B be the event that it is going to rain. We have the following information:

P(B) = 0.45

P(no rain) = P(B′) = 1 − P(B) = 1 − 0.45 = 0.55

The conditional probabilities of these events will be:

P(A|B) = 0.64

P(A|B') = 0.25

As the events B and B' form the partitions of the sample space, S, so, by the total probability theorem, we will compute the probability like this:

P(A) = P(B) P(A|B) + P(B′) P(A|B′)

= 0.45 x 0.64 + 0.55 x 0.25

= 0.4255

 

Example 3

Two dice are rolled simultaneously. Calculate:

a) The probability of two dice landing on the multiples of 3

b)  The probability that the sum of the numbers on two dice is equal to or greater than 11

c)  The probability that the sum of two numbers on the dice is less than 4

Solution

First, we will find the sample space. The sample space is the list of all probable results of a random experiment. The sample space of the two dice is given below:

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)}

The total number of possible outcomes is 36.

 

Part a

Here, we will compute the probability that two dice land on the multiples of 3. To find this probability, first, let us see the number of possible outcomes when two dice land on the multiples of 3:

{(3,3)(3,6), (6,3),(6,6)}

Number of possible outcomes = 4

Total number of possible outcomes  = 36

Probability of two dice landing on the odd numbers = \frac{4}{36} = \frac{1}{9}

 

Part b

To calculate the probability that the sum of the numbers on two dice is equal to or greater than 11, first, we will see the number of possible outcomes.

(5,6), (6,5), (6,6)

Number of possible outcomes = 3

Total number of possible outcomes = 36

Probability that the sum of the numbers on two dice is equal to or greater than 11 = \frac{3}{36} = \frac{1}{12}

 

Part c

Now, we will find the probability that the sum of two numbers on the dice is less than 4. For this, first, we should find the number of possible outcomes in which the sum of two numbers on the dice is less than 4.

    (1,1),(1,2), (2,1)

Number of possible outcomes = 3

Total number of possible outcomes = 36

The probability of having the sum of two numbers on the dice less than 4 = \frac{3}{36} = \frac{1}{12}

 

Example 4

In a class, 55% of the students passed a literature test, 40% passed a history test and 30% passed both literature and history tests. Suppose a student is chosen at random. If he/she has passed a literature test, what is the probability that he/she has also passed the history test?

Solution

Probability of passing a literature test = P(L) = 0.55

Probability of passing both literature and history tests = P(L \cap H) = 0.30

P (History|Literature) = \frac{P (L \cap H)} {P(L)}

= \frac {0.30} {0.55} = 0.54

 

Example 5

In a group, 45% of the people play football, 25% play basketball, and 15% play both basketball and football. Find the probability that a randomly chosen person plays a football, given that he already plays basketball.

Solution

Probability of randomly chosen person playing a football = P(F) = 0.45

Probability of randomly chosen person playing a football and a basketball = P(F \cap B)= 0.15

Probability that a randomly chosen person plays basketball = P(B) = 0.25

The probability that a randomly chosen person plays a football given that he already plays basketball = P(F|B) = \frac{P(F \cap B)}{P(B)} = \frac{0.15}{0.25} = 0.6

 

Example 6

A coin and a die are rolled simultaneously. What is the probability that the die will land on a prime number and coin on a head?

Solution

Total number of possible outcomes for a die = 6

Number of prime numbers on a die = 3

Probability of die landing on a prime number = \frac{1}{2}

Number of possible outcomes for a coin = 2

Probability that a coin will land on a head = \frac{1}{2}

The probability of die landing on a prime number and coin on a head = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

 

Example 7

Suppose five coins are tossed simultaneously. What is the probability of getting a head on the first coin, tail on the second coin, head on the third coin, tail on the fourth coin, and tail on the fifth coin?

Solution

Probability of getting head on the first coin = \frac{1}{2}

Probability of getting tail on the second coin = \frac{1}{2}

Probability of getting head on the third coin = \frac{1}{2}

Probability of getting tail on the fourth coin = \frac{1}{2}

Probability of getting tail on the fifth coin = \frac{1}{2}

The probability of getting a head on the first coin, tail on the second coin, head on the third coin, tail on the fourth coin, and tail on the fifth coin = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} =\frac{1}{32}$

 

 

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Rafia Shabbir