# A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box.

Find the probability that number on the drawn card is

(i) divisible by 2 or 3

(ii) a prime number

Last updated date: 28th Mar 2023

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Answer

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312k+ views

Hint: Use Probability=favorable cases/total cases.

(i) Let $E$ be the event of drawing a card with number divisible by 2 or 3 from the cards with

number 1 to 20

Number divisible by 2 or 3 from 1 to 20 = 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20

No. of favorable outcomes = 13

Total no. of possible outcomes = 20

We know that, Probability $P\left( E \right) = \dfrac{{\left( {{\text{No}}{\text{. of favorable outcomes}}} \right)}}{{\left( {{\text{Total no}}{\text{. of possible outcomes}}} \right)}} = \dfrac{{13}}{{20}}$

Therefore, the probability that a number on the card selected from numbers 1 to 20 divisible

by 2 and 3 is $\dfrac{{13}}{{20}}$

(ii)

Let $F$ be the event of drawing a card with a prime number from the cards with numbers 1 to 20. Prime numbers are those which are divisible by 1 and the number itself.

Prime number from 1 to 20 = 2, 3, 5, 7, 11, 13, 17, 19

No. of favorable outcomes = 8

Total no. of possible outcomes = 20

We know that, Probability $P\left( F \right) = \dfrac{{\left( {{\text{No}}{\text{. of favorable outcomes}}} \right)}}{{\left( {{\text{Total no}}{\text{. of possible outcomes}}} \right)}} = \dfrac{8}{{20}} = \dfrac{2}{5}$

Therefore, the probability that a card with prime number is selected from numbers 1 to 20 $ = \dfrac{2}{5}$

Note: We can solve the above question by creating sample space as well as using combinations. Using

combination theorems will be much easier as compared to creating sample space. Mutually exclusive events need to be kept in mind.

(i) Let $E$ be the event of drawing a card with number divisible by 2 or 3 from the cards with

number 1 to 20

Number divisible by 2 or 3 from 1 to 20 = 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20

No. of favorable outcomes = 13

Total no. of possible outcomes = 20

We know that, Probability $P\left( E \right) = \dfrac{{\left( {{\text{No}}{\text{. of favorable outcomes}}} \right)}}{{\left( {{\text{Total no}}{\text{. of possible outcomes}}} \right)}} = \dfrac{{13}}{{20}}$

Therefore, the probability that a number on the card selected from numbers 1 to 20 divisible

by 2 and 3 is $\dfrac{{13}}{{20}}$

(ii)

Let $F$ be the event of drawing a card with a prime number from the cards with numbers 1 to 20. Prime numbers are those which are divisible by 1 and the number itself.

Prime number from 1 to 20 = 2, 3, 5, 7, 11, 13, 17, 19

No. of favorable outcomes = 8

Total no. of possible outcomes = 20

We know that, Probability $P\left( F \right) = \dfrac{{\left( {{\text{No}}{\text{. of favorable outcomes}}} \right)}}{{\left( {{\text{Total no}}{\text{. of possible outcomes}}} \right)}} = \dfrac{8}{{20}} = \dfrac{2}{5}$

Therefore, the probability that a card with prime number is selected from numbers 1 to 20 $ = \dfrac{2}{5}$

Note: We can solve the above question by creating sample space as well as using combinations. Using

combination theorems will be much easier as compared to creating sample space. Mutually exclusive events need to be kept in mind.

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