Question

Six digits are selected at random again and again from a random number table and the even digits are

counted each time. In most of the cases, the number of even digits will be 3 how?
​

Answer 1

Given:

Any six digits are selected at random from a number table.

Even digits are counted every time.

To Find:

Proof to say that the even digits will be e in most of the cases.

Solution:

We can simply solve this numerical problem by using the following process.

Number of odd digits = 1,3,5,7,9 = 5

Number of even digits = 0,2,4,6,8 = 5

Total number of digits = 10

Now, the probability of picking even digits = ( no.of even digits )/ ( Total no.of digits )

⇒ Probability of picking even digits = 5/10

⇒ Probability of picking even digits = 1/2

Hence, out of 6 digits, 3 will be even in most of the cases.

#SPJ2

Answer 2

Explanation:

Given that a random number is to be drawn from a random number table.

The numbers which we have been  $0,1,2,3,4,5,6,7,8,9$.

Out of these, even numbers  ⇒  $0,2,4,6,8$

                      Odd numbers   ⇒ $1,3,5,7,9$

So the probability of getting  an even number

                                               = ( No. of even digits ) / ( No. of total digit )

                                                =  $\frac{5}{10}$  

                                                $=\frac{1}{2}$

Therefore, the chances of getting an even number

out of $6$ digits  = $3$.

#SPJ3