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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

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Last updated date: 13th Jul 2024
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Answer
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453.9k+ 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.