Question

Tickets numbered 1 to 25 are mixed up and a ticket
is drawn at random. What is the probability that the
ticket drawn has a number which is a multiple of both
2 and 3 ?​

Answer 1

Answer:

$\large\boxed{\sf{0.16}}$

Step-by-step explanation:

Given that,

Tickets numbered 1 to 25 are mixed up.

Now, one ticket is chosen at random.

To find the probability that the ticket drawn has a number which is multiple of both 2 and 3.

Now, we know that, if a number is multiple of both 2 and 3, then it's the multiple of 6.

Therefore, we have,

Favourable outcomes = 6, 12, 18, 24

Now, clearly, we have,

Total number of possible outcomes = 25

No. of favourable outcomes = 4

Also, we know that,

Probability (P) is the ratio of number of favourable outcomes to number of total possible outcomes.

Therefore, we will get,

=> P = 4/25

=> P = (4 × 4)/ (25 × 4)

=> P = 16/100

=> P = 0.16

Hence, the required probability is 0.16.

Answer 2

$\huge{ \mathtt{ \fbox{Solution :) }}}</strong><strong> </strong><strong> </strong><strong>$

Given ,

• No. of total outcomes = 25 , because the total numbers of tickets is 25

• No. of favourable outcomes = 4 because 6 , 12 , 18 and 24 ticket numbers are multiples of 2 and 3

We know that ,

$\large \mathtt{\fbox{Probability \: (P) = \frac{No. \: of \: favorable \: outcomes}{No. \: of \: total \: outcomes} }}$

Thus ,

P = 4/25

P = 0.16

Hence , the required probability is 0 16

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