Question

Ajay starts a circus. But due to small size of his circus he can accomodate only 100 viewers. His 100 circus
passes i.e. tickets are numbered as 1,2,3,4,5,6,... so on up to 100. During one of the shows, he
decided to give a gift to only one viewer. What is the probability that he chooses a viewer who has the
digit 2 printed on his/her pass?
19/100
$ 21/100
32/100
1/5​

Answer 1

Answer:

The probability that he chooses a viewer who has digit 2 printed on his/her pass is 19/100.

Step-by-step explanation:

Given,

Ajay numbers the tickets as 1, 2, 3, 4...100.

To find,

The probability of winning a gift to a viewer who has digit 2 printed on his/her ticket.

Calculation,

Let S be the sample space

So, n(S) = 100 (As there are 100 participants)

Let E be the event of winning the gift to a viewer who has digit 2 printed on his/her ticket.

Then, E = 2, 12, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 32, 42, 52, 62, 72, 82, 92.

So, n(E) = 19

Hence, p(E) = n(E)/n(S)

p(E) = 19/100

Therefore, the probability that he chooses a viewer who has digit 2 printed on his/her pass is 19/100.

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Answer 2

Answer:

The probability of choosing a ticket from 1 to 100  with '2'  printed on it = $\frac{19}{100}$

Step-by-step explanation:

Given,

The numbers in the 100 circus tickets are 1,2,3,4,..................,100

To find,

The probability of getting a digit '2' in the ticket.

Solution:

Recall the concept

Probability of an event = $\frac{Number \ of \ favorable \ outcomes }{Total \ number \ of \ outcomes}$

Here, the total number of outcomes =  The numbers from 1 to 100 = 100

Number of favorable outcomes = The numbers from 1 to 100 with digits 2 in it

Favourable outcomes from 1 to 100 with 2 printed = 2,12, 20,21,22,23,24,25,26,27,28,29,32,42,52,62,72,82,92

Number of favorable outcomes = 19

Hence the probability of choosing a ticket from 1 to 100  with '2'  printed on it = $\frac{19}{100}$

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