Answer
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Hint: Probability is a measure of how likely an event is to occur. An event is a number between \[0\] and \[1,\] where \[0\] indicates impossibility of the event and $1$ indicates certainty of the event. The formula used to calculate probability is:
Probability of an event P(E) $ = \dfrac{{{\text{favourable outcomes}}}}{{{\text{total outcomes}}}}$
To solve this question, i.e., to know the probability of getting an odd number at least once, we need to obtain the probability of getting even number three times and then subtract it from \[1.\]
where, \[1\] indicates the certainty of the event.
Complete step-by-step solution
Step 1: We need to throw a die, so the total outcomes we will get of throwing die one time \[ = {\text{ }}\left\{ {1,2,3,4,5,6} \right\}{\text{ }} = {\text{ }}6\]
And the favourable outcome, i.e., number of odd numbers we will get \[ = {\text{ }}\left\{ {1,3,5} \right\}{\text{ }} = {\text{ }}3\]
Similarly, number of even numbers we will get \[ = {\text{ }}\left\{ {2,4,6} \right\}{\text{ }} = {\text{ }}3\]
Step 2: So, the probability of getting an odd number at first throw \[ = \] $\dfrac{3}{6} = \dfrac{1}{2}$
Similarly, the probability of getting an even number at first throw \[ = \] $\dfrac{3}{6} = \dfrac{1}{2}$
Step 3: So, the probability of getting an even number three times $ = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}$
Step 4: We need to find probability of getting an odd number at least once, so for that we will use the formula mentioned below,
Probability of getting an odd number at least once \[ = \] \[1{\text{ }}-\] Probability of never getting an odd number in none of the throws (or probability of getting even number three times)
$ = 1 - \dfrac{1}{8}$
$ = \dfrac{7}{8}$
Therefore, the probability of getting an odd number at least once is $\dfrac{7}{8}$
Note:To solve the question, we need to make a set for total number of outcomes and a set of number of favourable outcomes. And then calculating three times, as according to the question die is rolled thrice.
Probability of an event P(E) $ = \dfrac{{{\text{favourable outcomes}}}}{{{\text{total outcomes}}}}$
To solve this question, i.e., to know the probability of getting an odd number at least once, we need to obtain the probability of getting even number three times and then subtract it from \[1.\]
where, \[1\] indicates the certainty of the event.
Complete step-by-step solution
Step 1: We need to throw a die, so the total outcomes we will get of throwing die one time \[ = {\text{ }}\left\{ {1,2,3,4,5,6} \right\}{\text{ }} = {\text{ }}6\]
And the favourable outcome, i.e., number of odd numbers we will get \[ = {\text{ }}\left\{ {1,3,5} \right\}{\text{ }} = {\text{ }}3\]
Similarly, number of even numbers we will get \[ = {\text{ }}\left\{ {2,4,6} \right\}{\text{ }} = {\text{ }}3\]
Step 2: So, the probability of getting an odd number at first throw \[ = \] $\dfrac{3}{6} = \dfrac{1}{2}$
Similarly, the probability of getting an even number at first throw \[ = \] $\dfrac{3}{6} = \dfrac{1}{2}$
Step 3: So, the probability of getting an even number three times $ = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}$
Step 4: We need to find probability of getting an odd number at least once, so for that we will use the formula mentioned below,
Probability of getting an odd number at least once \[ = \] \[1{\text{ }}-\] Probability of never getting an odd number in none of the throws (or probability of getting even number three times)
$ = 1 - \dfrac{1}{8}$
$ = \dfrac{7}{8}$
Therefore, the probability of getting an odd number at least once is $\dfrac{7}{8}$
Note:To solve the question, we need to make a set for total number of outcomes and a set of number of favourable outcomes. And then calculating three times, as according to the question die is rolled thrice.
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