Question

A 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?

Answer 1

Answer:    3/14

Step-by-step explanation:

Total balls = 21

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

Then P(A) = (even balls ) / (total 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

Answer 2

Given: A box contains 21 balls numbered 1 to 21. A ball is drawn and then another ball is drawn without replacement.

To find: The probability that both balls are even numbered.

Solution:

When the first ball is drawn, the number of even-numbered balls is 10 and the total number of balls is 21. Thus, the probability of getting an even-numbered ball in the first draw can be written as follows.

$P(1) = \frac{10}{21}$

According to the question, no replacement takes place before a ball is drawn the second time. So, the number of even-numbered balls left must be 9, given that the first ball drawn was even-numbered. Similarly, the total number of balls must be 20.

$P(2) = \frac{9}{20}$

To calculate the probability that both the balls are even-numbered, the following calculation is done.

$P(E) = P(1)*P(2)$

         $= \frac{10}{21} *\frac{9}{20}$

         $= \frac{3}{14}$

Therefore, the probability that both balls are even-numbered is 3/14.