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


_________________

I'm the Dumbest of All !!

METHOD-1

No. of ways for Ann to select a ball out of 10 balls = 10C1 = 10 ways
No. of ways for Jane to select Same ball that ann select = 1C1 = 1 ways

Total Favourable ways of Ann and Jane to pick the same ball = 10*1 = 10 ways

Total ways of Ann and Jane to pick balls randomly = 10*10 = 100 ways

Probability = Favourable Outcomes / Total Outcomes = 10/100 = 1/10
Answer: option D

METHOD-2

Probability of Ann to select any ball out of 10 balls = 10/10 = 1
Probability of Jane to select Same ball that Ann selected out of a total 10 balls = 1/10

Total Probability of Both events to happen = 1*(1/10) = 1/10
Answer: option D
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

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
_________________