Question

A number is selected at random from first 20 natural numbers . Find the

probability of the

number being: (2)

(a) a prime number (b) a multiple of 6 (c) an even number

(d) numbers which are divisible both by 4 and 6​

Answer 1

Answer:

a) 8/20=2/5

b)3/20

c)10/20=1/2

d)1/20

Answer 2

Given :

A number is selected at random from first 20 natural numbers.

To Find :

• a prime number
• a multiple of 6
• an even number
• numbers which are divisible both by 4 and 6

Solution :

Analysis :

Here the concept of probability is used. We first have to find the total number of outcomes. Then the favourable outcomes. After that by dividing both we will find it.

Explanation :

a) A prime number :

• Total number of outcomes = 20
• Favourable outcomes = 2, 3, 5, 7, 11, 13, 17, 19 = 8

So,

$\\ \bf P_{(prime\ number)}=\dfrac{Favourable\ Outcomes}{Total\ Outcomes}$

Substituting the values,

$\\ \implies\sf P_{(prime\ number)}=\dfrac{\not{8}\ \ ^2}{\cancel{20}\ \ ^5}$

$\\ \therefore\boxed{\bf P_{(prime\ number)}=\dfrac{2}{5}}$

________________________

b) Multiple of 6 :

• Total number of outcomes = 20
• Favourable outcomes = 6, 12, 18 = 3

So,

$\\ \bf P_{(multiple\ of\ 6)}=\dfrac{Favourable\ Outcomes}{Total\ Outcomes}$

Substituting the values,

$\\ \therefore\boxed{\bf P_{(multiple\ of\ 6)}=\dfrac{3}{20}}$

________________________

c) An even number :

• Total number of outcomes = 20
• Favourable outcomes = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 = 10

So,

$\\ \bf P_{(even\ number)}=\dfrac{Favourable\ Outcomes}{Total\ Outcomes}$

Substituting the values,

$\\ \implies\sf P_{(even\ number)}=\dfrac{1\!\!\!\!\not{0}}{2\!\!\!\!\not{0}}$

$\\ \therefore\boxed{\bf P_{(even\ number)}=\dfrac{1}{2}}$

________________________

d) Numbers divisible both by 4 & 6 :

• Total number of outcomes = 20
• Favourable outcomes = 12 = 1

So,

$\\ \bf P_{(numbers\ divisible\ by\ both\ 4\ and\ 6)}=\dfrac{Favourable\ Outcomes}{Total\ Outcomes}$

Substituting the values,

$\\ \therefore\boxed{\bf P_{(numbers\ divisible\ by\ both\ 4\ and\ 6)}=\dfrac{1}{20}.}$