Question

3️⃣A box contains 21 balls numbered 1 to 21. A ball is drawn and then another ball is drawn without replacement. What is the probability that both balls are even numbered?

A.27
B.821
C.314
D.521
E.None of these​

Answer 1

Answer:

3 / 14 option e is correct

Step-by-step explanation:

Let A be the event that the first ball is even.

Then P(A) = (# even balls ) / (# balls) = 10 / 21

Let B be the event that the second ball is even.

Then the conditional probability that B occurs given that A has occurred is:

P(B | A) = (# even balls left) / ( # balls left ) = 9 / 20

Finally,

P(both balls are even)

= P( A ∩ B )

= P(B | A) × P(A)

= (9/20) × (10/21)

= 3 / 14