Two fair dice are tossed. Let A be the event that the first die shows an even number and B be the event that the second die shows an odd number. The two events A and B are
A.Mutually exclusive
B.Independent and mutually exclusive
C.Dependent
D.None of the above
Answer
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Hint: Probability is defined as the events which are likely to happen, or probable and can be calculated by taking the ratio of the favorable outcomes with the total number of observations and then simplifying the ratio for the required value.
Complete step-by-step answer:
When two dices are tossed together, the possible outcomes will be as below –
\[
(1,1)\;{\text{(1,2) (1,3) (1,4) (1,5) (1,6)}} \\
(2,1)\;{\text{(2,2) (2,3) (2,4) (2,5) (2,6)}} \\
(3,1)\;{\text{(3,2) (3,3) (3,4) (3,5) (3,6)}} \\
(4,1)\;{\text{(4,2) (4,3) (4,4) (4,5) (4,6)}} \\
(5,1)\;{\text{(5,2) (5,3) (5,4) (5,5) (5,6)}} \\
(6,1)\;{\text{(6,2) (6,3) (6,4) (6,5) (6,6)}} \;
\]
When two dice are tossed together the total number of possible outcomes becomes $ = 36$
By observation in the total possible outcomes we can find -
The probability that the first die shows an even number can be given as $P(A) = \dfrac{{18}}{{36}} = \dfrac{1}{2}$
Similarly, the probability that the second die shows an odd number can be given as $P(B) = \dfrac{{18}}{{36}} = \dfrac{1}{2}$
The probability that the first die shows even and the second die shows an odd number can be given as $P(A \cap B) = \dfrac{9}{{36}} = \dfrac{1}{4}$ …. (A)
Hence, they are not mutually exclusive, $P(A).P(B) = \dfrac{1}{2} \times \dfrac{1}{2}$
Simplify finding the product of the terms –
$P(A).P(B) = \dfrac{1}{4}$ …(B)
From (A) and (B)
$P(A \cap B) = P(A).P(B)$
Hence, both the given events are independent.
Therefore, from the given multiple choices – the option D is the correct answer.
So, the correct answer is “Option B”.
Note: Know the difference between the types of events. Mutually exclusive events are defined as the mutually exclusive events where two events can not occur at the same time whereas in the independent events, the first event occurs without affecting the second event.
Complete step-by-step answer:
When two dices are tossed together, the possible outcomes will be as below –
\[
(1,1)\;{\text{(1,2) (1,3) (1,4) (1,5) (1,6)}} \\
(2,1)\;{\text{(2,2) (2,3) (2,4) (2,5) (2,6)}} \\
(3,1)\;{\text{(3,2) (3,3) (3,4) (3,5) (3,6)}} \\
(4,1)\;{\text{(4,2) (4,3) (4,4) (4,5) (4,6)}} \\
(5,1)\;{\text{(5,2) (5,3) (5,4) (5,5) (5,6)}} \\
(6,1)\;{\text{(6,2) (6,3) (6,4) (6,5) (6,6)}} \;
\]
When two dice are tossed together the total number of possible outcomes becomes $ = 36$
By observation in the total possible outcomes we can find -
The probability that the first die shows an even number can be given as $P(A) = \dfrac{{18}}{{36}} = \dfrac{1}{2}$
Similarly, the probability that the second die shows an odd number can be given as $P(B) = \dfrac{{18}}{{36}} = \dfrac{1}{2}$
The probability that the first die shows even and the second die shows an odd number can be given as $P(A \cap B) = \dfrac{9}{{36}} = \dfrac{1}{4}$ …. (A)
Hence, they are not mutually exclusive, $P(A).P(B) = \dfrac{1}{2} \times \dfrac{1}{2}$
Simplify finding the product of the terms –
$P(A).P(B) = \dfrac{1}{4}$ …(B)
From (A) and (B)
$P(A \cap B) = P(A).P(B)$
Hence, both the given events are independent.
Therefore, from the given multiple choices – the option D is the correct answer.
So, the correct answer is “Option B”.
Note: Know the difference between the types of events. Mutually exclusive events are defined as the mutually exclusive events where two events can not occur at the same time whereas in the independent events, the first event occurs without affecting the second event.
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