In this article, you will find detailed answers to the probability questions. So, let us get started.

Example 1

Sam wants to go to the doctor for a regular checkup. The probabilities of visiting a doctor with or without traffic are 0.33 and 0.74 respectively. If the probability of traffic on a certain day is 0.53, then determine the probability that John will go to the doctor?

Solution

Suppose A be the event that John will go to the doctor and B be the event that there is traffic. We have the following information:

P(B) = 0.53

P(no traffic) = P(B′) = 1 − P(B) = 1 − 0.53 = 0.47

The conditional probabilities of these events will be:

P(A|B) = 0.33

P(A|B') = 0.74

We know that the events B and B' form the partitions of the sample space, S, so, using the total probability theorem, we will compute the probability like this:

P(A) = P(B) P(A|B) + P(B′) P(A|B′)

= 0.53 x 0.33 + 0.47 x 0.74

= 0.5227

 

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Example 2

Suppose three dice are rolled simultaneously. What is the probability of getting a multiple of 2 on the first die, a composite number on the second die, and 3 or 5 on the third die?

Solution

Total number of possible outcomes on a single die = 6

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

Probability of getting a composite number on the second die = \frac{2}{6} = \frac{1}{3}

Probability of getting 3 or 5 on the  third die = \frac{2}{6} = \frac{1}{2}

The probability of getting a multiple of 2 on the first die, a composite number on the second die, and 3 or 5 on the third die = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{12}

 

Example 4

In a class, 25% of the students play badminton, 40% play tennis, and 10% play both tennis and badminton. Suppose a student is chosen at random. If he/she plays tennis, what is the probability that he/she has also plays badminton?

Solution

Probability of the student playing a tennis = P(T) = 0.40

Probability of playing both badminton and tennis = P(B \cap T) = 0.10

P (Badminton|Tennis) = \frac{P (B \cap T)} {P(T)}

= \frac {0.10} {0.40} = 0.25

 

Example 5

In a class of 40 students, 30 have a laptop, 20 have a tablet, and 10 students have a laptop and tablet both. Find the probability that a randomly selected student:

a) have a laptop only

b) have a tablet only

c) have a laptop

d) have tablet

e) have tablet and laptop

f) have laptop given that he/she already owns a tablet

g) have a tablet given that he/she already owns a laptop

 

Solution

This example has several parts, so we will solve each part one by one. But before proceeding to solve the parts, first, we will draw a Venn diagram using the information from the examples like this:

Part a

Total number of students in a class = 40

Number of students who own a laptop only = 20

Probability of randomly selected student owning a laptop only = \frac{1}{2}

 

Part b

Total number of students in a class = 40

Number of students who own a tablet only = 10

Probability of randomly selected student owning a tablet only = \frac{1}{4}

 

Part c

Total number of students in a class = 40

Number of students who own a laptop = 30

Probability of randomly selected student owning a laptop = \frac{3}{4}

 

Part d

Total number of students in a class = 40

Number of students who own a tablet = 20

Probability of randomly selected student owning a tablet = \frac{1}{2}

 

Part e

Total number of students in a class = 40

Number of students who own a laptop and a tablet = 10

Probability of randomly selected student owning a laptop and a tablet = \frac{1}{4}

 

Part f

Probability of randomly selected student owning a laptop and a tablet = P(L \cap T) =\frac{1}{4}

Probability of randomly selected student owning a tablet = \frac{1}{2}

Probability of randomly selected student having laptop given that he/she already own a tablet = P(L|T) = \frac{P(L \cap T)} {P(T)} = \frac{\frac{1}{4}} {\frac{1}{2}} = \frac{1}{2}

 

Part g

Probability of randomly selected student owning a laptop and a tablet = P(L \cap T) =\frac{1}{4}

Probability of randomly selected student owning a laptop = \frac{3}{4}

Probability of randomly selected student having a tablet given that he/she already owns a laptop = P(T|L) = \frac{P(L \cap T)} {P(L)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}

 

Example 6

In a box, there are 15 almonds, 18 cashew nuts, 25 peanuts, and 12 pistachios. What is the probability of:

a) drawing an almond randomly

b) drawing two cashew nuts

c) drawing a pistachio and then an almond without replacement

d) drawing an almond and then a peanut without replacement

e) drawing a dry fruit that is not pistachio

f) drawing an almond or a pistachio

Solution

Part a

Total number of dry fruits in a box = 15 + 18 + 12 + 25 = 70

Number of almonds in a box = 15

Probability of drawing an almond = \frac{15}{70} = \frac{3}{14}

 

Part b

Total number of dry fruits in a box =  70

Number of cashew nuts = 18

Probability of drawing a single cashew nut = \frac{18}{70} = \frac{9}{35}

Probability of drawing a cashew nut again = \frac{17}{69}

The probability of drawing two cashew nuts = \frac{9}{35} \cdot \frac{17}{69} = \frac{51}{805}

 

Part c

Total number of dry fruits in a box =  70

Number of pistachios in a box = 12

The probability of drawing a pistachio from the box = \frac{12}{70}

Number of almonds in a box = 15

Probability of drawing an almond from the box = \frac{15}{69}

Probability of drawing pistachio and then an almond from a box = \frac{12}{70} \times \frac{15}{69} = \frac{6}{161}

 

Part d

Total number of dry fruits in a box =  70

Number of almonds in a box = 15

Probability of drawing an almond from the box = \frac{15}{70}

Number of peanuts in a box = 25

Probability of drawing a peanut from the box after almond = \frac{25}{69}

The probability of drawing an almond and then a peanut from the box = \frac{15}{70} \times \frac{25}{69} = \frac{25}{322}

 

Part e

Total number of dry fruits in a box =  70

Number of pistachio in the box = 12

Number of dry fruits which aren't pistachios = 70 - 12 = 58

The probability of drawing a dry fruit which isn't a pistachio = \frac{58}{70} = \frac{29}{35}

 

Part f

Total number of dry fruits in a box =  70

Number of almonds in the box = 15

Number of pistachios in the box = 12

Probability of drawing an almond or a pistachio from the box = \frac{27}{70}

 

Example 7

50% of the residents in a certain area own a house. 35% of the residents live on rent, while 10% own a house and live on rent. For a randomly selected person, calculate the probability that the person owns a home, given that he lives on rent.

Solution

Probability of randomly selected person owning a house = 0.50

Probability of randomly selected person living on rent = 0.35

Probability of randomly selected person owning a house and living on rent = 0.10

The probability of randomly selected student owning a house, given that he is living on rent = P(H|R) = \frac{P(H \cap R)} {P(R)} = \frac{0.10}{0.35} = 0.28

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Rafia Shabbir