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$

{\text{A}}{\text{. }}\dfrac{6}{{20}} \\

{\text{B}}{\text{. }}\dfrac{7}{{20}} \\

{\text{C}}{\text{. }}\dfrac{8}{{20}} \\

{\text{D}}{\text{. }}\dfrac{9}{{20}} \\

$

Answer

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Hint- In this question we will firstly find the count of numbers which are divisible individually by 3 and 5. And then also find the numbers which are divisible by both then using the formula $n(A \cup B) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$ we will easily find the number of favourable outcomes.

__Complete step-by-step answer:__

Here we have been given a bag containing 20 tickets numbered from 1 to 20. Now, to find that the probability that the number drawn is divisible by 3 or 5.

So, let A be the event that the drawn number is divisible by 3

So, A = {3, 6, 9, 12, 15, 18}

$ \Rightarrow n\left( A \right) = 6$…………………….. Equation (1)

Now, let B be the event that the drawn number is divisible by 5

So, B= {5, 10, 15, 20}

$ \Rightarrow n\left( B \right) = 4$…………………….. Equation (2)

As we need to find the probability that the number drawn is divisible by 3 or 5.

So, it means we need to find $P(A \cup B)$ which is the number of favourable outcomes, for which we need to find $n(A \cup B)$.

And we know that $n(A \cup B) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$

So, we also need to find $n\left( {A \cap B} \right)$

As $A \cap B$ means the numbers divisible by 3 and 5

So, $A \cap B$ = {15}

$ \Rightarrow n\left( {A \cap B} \right) = 1$…………………….. Equation (3)

Using Equation (1), Equation (2) and Equation (2) in $n(A \cup B) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$

We get,

$n(A \cup B) = 6 + 4 - 1 = 9$…………… Equation (4)

Now, the sample space is given to be 20.

So, as the formula of probability is ${\text{Probability = }}\dfrac{{{\text{Number of Favourable Outcomes}}}}{{{\text{Sample Space}}}}$

Hence, the required probability is $ = \dfrac{9}{{20}}$

Hence the correct option is D.

Note- Whenever we face such types of problems the main point to remember is that we need to have a good grasp over probability. In this particular problem, students usually forget that they have included the numbers divisible by both numbers twice and give the wrong answer, hence using the formula of sets is the best way to avoid errors. So, here we find the union of both events to get the number of favourable outcomes as 9 and as we were given that the given numbers are from 1 to 20, which will help us get the required probability.

Here we have been given a bag containing 20 tickets numbered from 1 to 20. Now, to find that the probability that the number drawn is divisible by 3 or 5.

So, let A be the event that the drawn number is divisible by 3

So, A = {3, 6, 9, 12, 15, 18}

$ \Rightarrow n\left( A \right) = 6$…………………….. Equation (1)

Now, let B be the event that the drawn number is divisible by 5

So, B= {5, 10, 15, 20}

$ \Rightarrow n\left( B \right) = 4$…………………….. Equation (2)

As we need to find the probability that the number drawn is divisible by 3 or 5.

So, it means we need to find $P(A \cup B)$ which is the number of favourable outcomes, for which we need to find $n(A \cup B)$.

And we know that $n(A \cup B) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$

So, we also need to find $n\left( {A \cap B} \right)$

As $A \cap B$ means the numbers divisible by 3 and 5

So, $A \cap B$ = {15}

$ \Rightarrow n\left( {A \cap B} \right) = 1$…………………….. Equation (3)

Using Equation (1), Equation (2) and Equation (2) in $n(A \cup B) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$

We get,

$n(A \cup B) = 6 + 4 - 1 = 9$…………… Equation (4)

Now, the sample space is given to be 20.

So, as the formula of probability is ${\text{Probability = }}\dfrac{{{\text{Number of Favourable Outcomes}}}}{{{\text{Sample Space}}}}$

Hence, the required probability is $ = \dfrac{9}{{20}}$

Hence the correct option is D.

Note- Whenever we face such types of problems the main point to remember is that we need to have a good grasp over probability. In this particular problem, students usually forget that they have included the numbers divisible by both numbers twice and give the wrong answer, hence using the formula of sets is the best way to avoid errors. So, here we find the union of both events to get the number of favourable outcomes as 9 and as we were given that the given numbers are from 1 to 20, which will help us get the required probability.

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