Question

Q. Among the digits 4,5,6,7,8,9 at first a digit is chosen at random and then a second digit is selected from the remaining digits
If each digit has the same probability of being selected then the probability that an odd digit is selected second time is:
(a)3/5
(b)1/5
(c)1/2
(d) 1/3​

Answer 1

Answer:

a)3/5

Step-by-step explanation:

to select 1 time =3/6=1/2

•°•to select 2nd time=3/5

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Answer 2

Answer:

(c)1/2

Step-by-step explanation:

Given that ,

Among the digits 4,5,6,7,8,9 at first a digit is chosen at random and then a second digit is selected from the remaining digits.

If each digit has the same probability of being selected then the probability that an odd digit is selected second time is:

(a)3/5

(b)1/5

(c)1/2

(d) 1/3​

So,

at first time the available digits = 4,5,6,7,8,9.

favourable outcomes = 3

Total outcomes = 6

∴ P(E) = $\frac{favourable ~outcomes}{Total ~outcomes}$

         = $\frac{3}{6}$

         = $\frac{1}{2}$

And probability of odd at second time

P(E') =$\frac{3}{5}$

Therefore, the probability odd digits selected second time =  (1/2) × (3/5)

= 1/2

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