In throwing a dice, let A be the event 'coming up of an odd number', B be the event 'coming up of an even number', C be the event 'coming up of a number $ \geqslant 4$ ' and D be the event 'coming up of a number $ < 3$ ‘, then
A. A and B are mutually exclusive and exhaustive .
B. A and C are mutually exclusive and exhaustive.
C. A , C and D form an exhaustive system.
D. B , C and D form an exhaustive system.
Answer
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Hint: In order to solve this question we need to solve this step by step by looking into the options given and solving them one by one by eliminating the options we can get our required answer but also we need to first know about the terms Mutually Exclusive and Exhaustive Events as they are given in the options and the concepts of these terms are used in this question . . The two events are mutually exclusive if they cannot both be true . A clear example is the set of outcomes of a single coin toss , which can result in their head or tail , but not both . And Mutually exhaustive events are a set of events where at least one of the events must occur . Keeping these two terms in mind we are going to solve the question .
Complete step-by-step answer:
The prerequisite of the above question is to know how to apply the concepts of Mutually Exclusive and Exhaustive Events by just applying their properties which are as follows =>
Probability of intersection of event A and event B = $P(A \cap B) = 0$. Also,
Probability of union of event A and event B = $P(A \cup B) = 1$.
Now , we will work on making the sample space as given in the question –
In a dice , there are six possible outcomes {1,2,3,4,5,6} . As per the conditions given we are considering =
Event A 'coming up of an odd number' has 3 possible outcomes = {1,3,5}
Event B 'coming up of an even number' has 3 possible outcomes = {2,4,6}
Event C 'coming up of a number $ \geqslant 4$’ has 3 possible outcomes = {4,5,6}
Event C 'coming up of a number $ < 3$ ‘, has 2 possible outcomes = {1,2}
A and B have no element in common, and together they form the universal set for this trial. Hence, A and B are mutually exclusive and exhaustive. Option A is correct.
A and C have 5 common to them. So they are not exclusive. Option B is incorrect.
A, C and D taken together form the universal set. Hence, they form an exhaustive system. Option C is correct.
B, C and D do not form an exhaustive system because neither of them has 3 in it. Option D is incorrect .
By combining all the individual answers , we can choose the correct options by making appropriate selection of options which implies that option A and option C are the correct options .
So, the correct answer is “Option A and C”.
Note: Read the mathematical statement given in the question carefully and try to understand and frame it mathematically .
The likelihood of an event to occur or the extent to which the event is probable is called probability . There are many more formulae which are used in different probability questions which arise due to different situations and events .
Ensure that you make the correct selection of options and not get confused .
Always remember that the likelihood of all events in a sample space adds up to 1 .
To check if your probability answer is correct or not , always remember that the likelihood of an event ranges from 0 to 1 , where 0 means the event to be an impossible one , and one represents the event to be certain .
Complete step-by-step answer:
The prerequisite of the above question is to know how to apply the concepts of Mutually Exclusive and Exhaustive Events by just applying their properties which are as follows =>
Probability of intersection of event A and event B = $P(A \cap B) = 0$. Also,
Probability of union of event A and event B = $P(A \cup B) = 1$.
Now , we will work on making the sample space as given in the question –
In a dice , there are six possible outcomes {1,2,3,4,5,6} . As per the conditions given we are considering =
Event A 'coming up of an odd number' has 3 possible outcomes = {1,3,5}
Event B 'coming up of an even number' has 3 possible outcomes = {2,4,6}
Event C 'coming up of a number $ \geqslant 4$’ has 3 possible outcomes = {4,5,6}
Event C 'coming up of a number $ < 3$ ‘, has 2 possible outcomes = {1,2}
A and B have no element in common, and together they form the universal set for this trial. Hence, A and B are mutually exclusive and exhaustive. Option A is correct.
A and C have 5 common to them. So they are not exclusive. Option B is incorrect.
A, C and D taken together form the universal set. Hence, they form an exhaustive system. Option C is correct.
B, C and D do not form an exhaustive system because neither of them has 3 in it. Option D is incorrect .
By combining all the individual answers , we can choose the correct options by making appropriate selection of options which implies that option A and option C are the correct options .
So, the correct answer is “Option A and C”.
Note: Read the mathematical statement given in the question carefully and try to understand and frame it mathematically .
The likelihood of an event to occur or the extent to which the event is probable is called probability . There are many more formulae which are used in different probability questions which arise due to different situations and events .
Ensure that you make the correct selection of options and not get confused .
Always remember that the likelihood of all events in a sample space adds up to 1 .
To check if your probability answer is correct or not , always remember that the likelihood of an event ranges from 0 to 1 , where 0 means the event to be an impossible one , and one represents the event to be certain .
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