Question

Mark the correct alternative in each of the following: A bag contains cards numbered from 1 to 25. A card is drawn at random from the bag. The probability that the number on this card is divisible by both 2 and 3 is
(a)$\frac{1}{5}$
(b)$\frac{3}{25}$
(c)$\frac{4}{25}$
(d)$\frac{2}{25}$

Answer 1

SOLUTION :

The correct option is (c)   : 4/25

Given : Cards numbered in a bag from 1 to 25

Total number of possible outcome = 25

Let E = Event of getting a number divisible by both 2 and 3

Number divisible by both 2 and 3 from 1 to 25 are = 6, 12, 18, 24

Number of outcomes  favorable to E = 4

Probability ,P(E) = Number of favourable outcomes / total number of outcomes

P(E) = 4/25

Hence, the Probability of getting a  number divisible by both 2 and 3 from 1 to  25 ,P(E)  is 4/25.

HOPE THIS ANSWER WILL HELP  YOU….

Answer 2

$\sf{Answer -}$

Option C is correct.

$\sf{Step-by-step\:explanation -}$

Given, a bag contains cards numbered from 1 to 25.

To find the probability that the number is divisible by 2 or 3.

From numbers 1-25, numbers divisible by 2 are -

$\implies 2,4,6,8,10,12,14,16,18,20,22,24$

From numbers 1-25, numbers divisible by 3 are -

$\implies 3,6,9,12,15,18,21,24$

From numbers 1-25, numbers divisible by both 2 and 3 are -

$\implies 6,12,18,24$

So, total number of favourable outcomes = $4$

Number of total outcomes = $25$

Probability -

$\implies \frac{Favourable\:outcome}{Total\:outcome}$

$\implies \frac{4}{25}$

Thus, option C is correct.