A box contains 10 balls numbered from 1 to 10 inclusive. If Ann remove

Question

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

A. 1/100
B. 1/90
C. 1/45
D. 1/10
E. 41/45

P(Ann selecting a ball) =  \frac{1}{10} 
P(jane selecting a ball) = \frac{1}{10} 
P(both removed the same ball) =  \frac{1}{10}  * \frac{1}{10} =  \frac{1}{100}

isn't correct ?

Bunuel · Accepted Answer

This would have been correct if it were "what's the probability that Ann and Jane removed some specific ball, let's say ball #3?".

Then we would have: the probability of Ann picking the ball #3 is 1/10 and the probability of Jane picking the same ball after the replacement is also 1/10 --> so, P(#3, #3)=1/10*1/10=1/100.

But since we are not asked about some specific ball then Ann can pick any ball, P=1 and the probability of Jane picking the same one would be 1/10 --> so, P(any, the one Ann picked)=1*1/10=1/10.

Or think about this way: we are basically asked about the probability of Jane picking some specific ball out of 10 which was previously marked by Ann, so it's simply 1/10.

Hope it's clear.

GMATinsight · Answer

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

HKD1710 · Answer

For Ann or For Jane TOTAL cases = 10 For Ann favorable cases = 10 For Ann probability of picking the ball = 10/10 = 1 Now that she has replaced the ball TOTAL cases remains the same. For Jane favorable case = 1 (If annhad picked the ball number 2, then jane has to pick only that ball, so only one choice for jane, hence only one fav case) Probability for Jane picking the same ball = 1/10. D is the right answer

ScottTargetTestPrep · Answer

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

BrentGMATPrepNow · Answer

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

MAMannan · Answer

Ways to choose for Ann = 10/10 = 1.
When this one is placed again in the box, the total number of balls in the box still remains 10.
So, Jane has only one chance to select a ball out of 10 balls.
Jane can choose = 1/10.
So the probability will be 1 x (1/10) = 1/10
Cheers!!

mauros · Answer

I made:

1 - (10/10 * 9/10)

You can see it by thinking 1 - P(not removing the same), Ann has 10/10 balls to remove, then Jane has only 9/10 to not remove the same as Ann

A box contains 10 balls numbered from 1 to 10 inclusive. If Ann remove

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A box contains 10 balls numbered from 1 to 10 inclusive. If Ann remove

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards