huda wrote:
A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and then returns to the box, and then Jane removes a ball at random, what is the probability that both women removed the same ball?
A. \(\frac{1}{100}\)
B. \(\frac{1}{90}\)
C. \(\frac{1}{45}\)
D. \(\frac{1}{10}\)
E. \(\frac{41}{45}\)
P(Ann and Jane remove same ball) = P(Ann removes ANY ball
AND Jane's ball matches Ann's ball)
= P(Ann removes ANY ball)
x P(Jane's ball matches Ann's ball)
= 1
x 1/10
= 1/10
Answer: D
Aside: Once Jane removes her ball (and then replaces it), we have 10 balls, and 1 of them is the one that Jane picked.
So, P(Jane's ball matches Ann's ball) = 1/10
Cheers,
Brent