A bag contains cards numbered from 1to 25.A card is drawn at random from the bag. Find the probability that the number on this card is (I) divisible by both 2 and 3 (ii) a two digit number?

Question

Answer

(1) - 4/25 = 0.16

(2) - 16/25 = 0.64

Answer

Number on cards are from 1 to 25,
(i) numbers between 1 to 25 divisible by 2 and 3 are ,
6,12,18,24
so,p( getting a number divisible by 2 or 3)=
number of events of favourable outcome/ total number of outcomes,
=4/25,
(ii) two digit number b/W [1,25]are ,
10,11,12,13,....25,
p(E)=16/25

Answer

4/25

Answer

1) 16/252)3/5

Answer

I) Number which is divisible by 2 and 3 both is divisible by 6 and there are 4 such numbers between 1 and 25, so the probability will be 4/25 = 0.16
ii) There are 16 two digit numbers between 1 and 25 so the probability will be 16/25 = 0.64

Answer

Divisible by both 2 and 3:
- Numbers divisible by both 2 and 3 are multiples of 6.
- The numbers in the given range that are multiples of 6 are 6, 12, 18, and 24.
- Therefore, there are 4 favorable outcomes.
- Total number of cards in the bag = 25.
- So, the probability of drawing a card divisible by both 2 and 3 is 4/25.

Two-digit numbers:
- Two-digit numbers in the given range are from 10 to 99.
- Within the range of 1 to 25, there are no two-digit numbers.
- Therefore, the probability of drawing a two-digit number is 0.

In conclusion, the probability of drawing a card that is divisible by both 2 and 3 is 4/25, and the probability of drawing a two-digit number is 0.

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