Question

The tickets in a box are numbered from 1 to 20 inclusive. If a ticket is drawn at random and replaced, and then a second ticket is drawn at random, what is the probability that the sum of their numbers is even? 1/2 1/4 1/5

Answer 1

What can happen with equal probability is:

even + even = even
even + odd = odd
odd + even = odd
odd + odd = even

So as much chance of even as for odd. The chance will be 1/2.

Answer 2

Answer: The correct option is (A)
`(1)/(2).`

Step-by-step explanation: Given that the tickets in a box are numbered from 1 to 20 inclusive. A a ticket is drawn at random and replaced, and then a second ticket is drawn at random.

We are to find the probability that the sum of their numbers is even.

The sum of the numbers of the two tickets will be even if either both the tickets have even number or both have odd numbers.

Out of 20 numbers, number of even numbers is 10, i.e., 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.

And, number of odd numbers is also 10, i.e., 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.

Case 1: The probability of drawing an even numbered ticket is given by

`p=(10)/(20)\\\\\\\Rightarrow p=(1)/(2).`

Since the first drawn ticket is replaced, so the probability of drawing an even numbered ticket again is given by

`p'=(10)/(20)\\\\\\\Rightarrow p'=(1)/(2).`

Therefore, the required probability of drawing an even numbered ticket both the times will be

`P_1=p* p'=(1)/(2)*(1)/(2)=(1)/(4).`

Case 2: The probability of drawing an odd numbered ticket is given by

`p=(10)/(20)\\\\\\\Rightarrow p=(1)/(2).`

Since the first drawn ticket is replaced, so the probability of drawing an odd numbered ticket again is given by

`p'=(10)/(20)\\\\\\\Rightarrow p'=(1)/(2).`

Therefore, the required probability of drawing an odd numbered ticket both the times will be

`P_2=p* p'=(1)/(2)*(1)/(2)=(1)/(4).`

Thus, the required probability that the sum of the numbers on both the tickets is even is

`P=P_1+P_2=(1)/(4)+(1)/(4)=(2)/(4)=(1)/(2).`

Option (A) is CORRECT.