Answer
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Hint: A die having 6 numbers total, then we will get 6 outcomes possible. First we will find the probability of getting a number less than 5 and then find the probability of getting an odd number and we will make Venn diagrams take integration of both of them.
Complete step-by-step answer:
We know that a die has 6 sides, and from the question we have to find the probability of getting a number less than 5 and an odd number in a single throw.
Now we using the formula of probability that is
${\text{p}}\left( {\text{A}} \right){\text{ = }}\dfrac{{{\text{number of favourable outcome}}}}{{{\text{total number of favourable outcome}}}}$
Now let A is the number less than 5, it is 4 because it can repeat.
So the probability of getting number less than 5 is
$p\left( A \right) = \dfrac{4}{6} = \dfrac{2}{3}$
(A is number of less than $5$)
And let B is the number in a single throw of a symmetrical die that is 3 because we cannot repeat the number and we have given in the question that the number is less than 5.
So the probability of getting number in single throw is
\[p\left( B \right) = \dfrac{3}{6} = \dfrac{1}{2}\]
(B is odd number)
Now we take an intersection from the number set of less than 5 and the odd number which cannot repeat.
So set of A is $\left( A \right) = \left\{ {1,2,3,4} \right\}$ and set of B is $\left( B \right) = \left\{ {1,2,3} \right\}$
And we know that the formula of intersection of two sets is
$
p\left( {A \ and B} \right) = p\left( A \right).p\left( B \right) \\
p\left( {A \ and B} \right) = \dfrac{2}{3} \times \dfrac{1}{2} \\
$
$p\left( {A \ and B} \right) = \dfrac{1}{3}$
Now using the formula of Venn diagram for A and B.
$p\left( {A \ or B} \right) = p\left( A \right) + p\left( B \right) - p\left( {A \ and B} \right)$
$
p\left( {A \ or B} \right) = \dfrac{2}{3} + \dfrac{1}{2} - \dfrac{1}{3} \\
p\left( {A \ or B} \right) = \dfrac{{4 + 3 - 2}}{6} \\
p\left( {A \ or B} \right) = \dfrac{5}{6} \\
$
So the required answer is $\dfrac{5}{6}$.
Note: These questions we solve by a simple probability method, first we will take probability for different conditions, after that binomial expression we will find probability of required statement.
Complete step-by-step answer:
We know that a die has 6 sides, and from the question we have to find the probability of getting a number less than 5 and an odd number in a single throw.
Now we using the formula of probability that is
${\text{p}}\left( {\text{A}} \right){\text{ = }}\dfrac{{{\text{number of favourable outcome}}}}{{{\text{total number of favourable outcome}}}}$
Now let A is the number less than 5, it is 4 because it can repeat.
So the probability of getting number less than 5 is
$p\left( A \right) = \dfrac{4}{6} = \dfrac{2}{3}$
(A is number of less than $5$)
And let B is the number in a single throw of a symmetrical die that is 3 because we cannot repeat the number and we have given in the question that the number is less than 5.
So the probability of getting number in single throw is
\[p\left( B \right) = \dfrac{3}{6} = \dfrac{1}{2}\]
(B is odd number)
Now we take an intersection from the number set of less than 5 and the odd number which cannot repeat.
So set of A is $\left( A \right) = \left\{ {1,2,3,4} \right\}$ and set of B is $\left( B \right) = \left\{ {1,2,3} \right\}$
And we know that the formula of intersection of two sets is
$
p\left( {A \ and B} \right) = p\left( A \right).p\left( B \right) \\
p\left( {A \ and B} \right) = \dfrac{2}{3} \times \dfrac{1}{2} \\
$
$p\left( {A \ and B} \right) = \dfrac{1}{3}$
Now using the formula of Venn diagram for A and B.
$p\left( {A \ or B} \right) = p\left( A \right) + p\left( B \right) - p\left( {A \ and B} \right)$
$
p\left( {A \ or B} \right) = \dfrac{2}{3} + \dfrac{1}{2} - \dfrac{1}{3} \\
p\left( {A \ or B} \right) = \dfrac{{4 + 3 - 2}}{6} \\
p\left( {A \ or B} \right) = \dfrac{5}{6} \\
$
So the required answer is $\dfrac{5}{6}$.
Note: These questions we solve by a simple probability method, first we will take probability for different conditions, after that binomial expression we will find probability of required statement.
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