Answer

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**Hint:**We start solving the problem by first writing the numbers on the cards that were divisible by both 2 and 7. We then count them to find the total number of cards that were satisfying the given condition. We then recall the definition of probability as $\text{probability = }\dfrac{\text{no}\text{. of ways to draw a card from favorable cards}}{\text{total no}\text{. of ways to draw a card from given 48 cards}}$ and use this along with the obtained number of cards to find the required probability. We similarly perform same steps for the remaining conditions to find the probability.

**Complete step by step answer:**

According to the problem, we are given that the cards marked with numbers 3, 4, 5, ……, 50 are placed within a box and mixed thoroughly and one card is drawn at random from the box.

(i) We need to find the probability that the number on the card is divisible by 2 and 7.

Let us write the set of numbers that were divisible both 2 and 7. Such numbers are $\left\{ 14,28,42 \right\}$.

So, the total number of terms that were divisible by both 2 and 7 from the given numbers is 3.

We know that the probability is defined as $\text{probability = }\dfrac{\text{no}\text{. of ways to draw a card from the 3 favorable cards}}{\text{total no}\text{. of ways to draw a card from given 48 cards}}$.

So, we have $p=\dfrac{3}{48}=\dfrac{1}{16}$.

∴ The probability of drawing the card that has a number on the card which is divisible by both 2 and 7 is $\dfrac{1}{16}$.

(ii) We need to find the probability that the number on the card is a perfect square.

Let us write the set of numbers that are perfect squares. Such numbers are $\left\{ 4,9,16,25,36,49 \right\}$.

So, the total number of terms that were perfect square from the given numbers is 6.

We know that the probability is defined as $\text{probability = }\dfrac{\text{no}\text{. of ways to draw a card from the 6 favorable cards}}{\text{total no}\text{. of ways to draw a card from given 48 cards}}$.

So, we have $p=\dfrac{6}{48}=\dfrac{1}{8}$.

∴ The probability of drawing the card that has a number on the card which is divisible by both 2 and 7 is $\dfrac{1}{8}$.

(iii) We need to find the probability that the number on the card is a perfect cube.

Let us write the set of numbers that were perfect cube. Such numbers are $\left\{ 8,27 \right\}$.

So, the total number of terms that were perfect cube from the given numbers is 2.

We know that the probability is defined as $\text{probability = }\dfrac{\text{no}\text{. of ways to draw a card from the 2 favorable cards}}{\text{total no}\text{. of ways to draw a card from given 48 cards}}$.

So, we have $p=\dfrac{2}{48}=\dfrac{1}{24}$.

∴ The probability of drawing the card that has the number on the card which is the perfect cube is $\dfrac{1}{24}$.

(iv) We need to find the probability that the number on the card is neither divisible by 3 nor by 5.

Let us write the set of numbers that were neither divisible by 3 nor by 5. Such numbers are $\left\{ 4,7,8,11,13,14,16,17,19,22,23,26,28,29,31,32,34,37,38,41,43,44,46,47,49 \right\}$.

So, the total number of terms that were neither divisible by 3 nor by 5 from the given numbers is 25.

We know that the probability is defined as $\text{probability = }\dfrac{\text{no}\text{. of ways to draw a card from the 25 favorable cards}}{\text{total no}\text{. of ways to draw a card from given 48 cards}}$.

So, we have $p=\dfrac{25}{48}$.

**∴ The probability of drawing the card that has a number on the card which is divisible by both 2 and 7 is $\dfrac{25}{16}$.**

**Note:**Whenever we get this type of problems, we first try to write the cases that were satisfying the given condition. We can also solve this problem by first finding the probability of the numbers that were not satisfying the given condition and then subtracting it from 1. We should know that the numbers should be present in the given numbers not from others. Similarly, we can expect problems to find the probability to draw a card that is divisible by 7.

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