shrive555 wrote:
A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball?
A. 1/100
B. 1/90
C. 1/45
D. 1/10
E. 41/45
Since Ann can pick any ball and Jane has only a 1/10 chance to match it, the probability that they select the same ball is:
1 x 1/10 = 1/10
Alternate Solution:
Many students think that the answer is 1/10 x 1/10 = 1/100, thinking that each woman has a 1/10 probability of picking a particular ball. Let’s look at the “long” solution to this problem to clarify the correct answer of 1/10.
Consider the ball marked “1.” The probability that Ann picks this ball is 1/10, and so the probability that Jane also picks this ball is 1/10. The probability that both women will pick the ball marked “1” is, therefore, 1/10 x 1/10 = 1/100.
Now consider the ball marked “2.” Again, the probability that Ann picks this ball is 1/10, and the probability that Jane also picks this ball is 1/10. Thus, the probability that both women will pick the ball marked “2” is, 1/10 x 1/10 = 1/100.
We continue this for the balls marked 3, 4, 5, 6, 7, 8, 9, and 10. For each ball, the probability will be 1/100.
Since there are 10 balls, the probability that Ann and Jane will pick the same ball is, therefore, 10 x 1/100 = 10/100 = 1/10.
Answer: D
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