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

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A box contains 10 balls numbered from 1 to 10 inclusive. If Ann remove  [#permalink]

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Updated on: 27 Jul 2015, 23:29
12
00:00

Difficulty:

25% (medium)

Question Stats:

60% (00:43) correct 40% (00:45) wrong based on 233 sessions

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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

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 ?

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Originally posted by shrive555 on 24 Nov 2010, 09:37.
Last edited by Bunuel on 27 Jul 2015, 23:29, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Posts: 50711
Re: A box contains 10 balls numbered from 1 to 10 inclusive. If Ann remove  [#permalink]

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24 Nov 2010, 09:48
8
6
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?

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 ?

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.
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A box contains 10 balls numbered from 1 to 10 inclusive. If Ann remove  [#permalink]

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28 Jul 2015, 07:29
3
1
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

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 ?

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

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
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Re: A box contains 10 balls numbered from 1 to 10 inclusive. If Ann remove  [#permalink]

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27 Apr 2016, 06:23
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

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 ?

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.
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Re: A box contains 10 balls numbered from 1 to 10 inclusive. If Ann remove  [#permalink]

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12 Nov 2018, 07:24
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.

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Re: A box contains 10 balls numbered from 1 to 10 inclusive. If Ann remove &nbs [#permalink] 12 Nov 2018, 07:24
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