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

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by BTGmoderatorDC » Fri Nov 09, 2018 12:04 am
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

OA D

Source: Magoosh
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A box contains 10 balls numbered from 1 to 10 inclusive. If

by fskilnik@GMATH » Fri Nov 09, 2018 5:13 am
BTGmoderatorDC 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
Source: Magoosh
$$?\,\, = \,\,P\left( {{\rm{same}}\,\,{\rm{ball}}\,\,{\rm{twice}}\,{\rm{,}}\,{\rm{with}}\,\,{\rm{replacement}}} \right)$$
Exactly 1 ball extracted in the second extraction (out of 10 equiprobable possibilities) will be the same as (any given) ball extracted in the first extraction.

Therefore the answer is 1/10, immediately.


Another approach:
$${\rm{Total}} = 10 \cdot 10\,\,{\rm{pairs}}\,\,\,\left( {{1^{{\rm{st}}}}\,,\,\,{2^{{\rm{nd}}}}} \right)\,{\rm{ extracted}}\,,\,\,{\rm{all}}\,\,{\rm{equiprobable}}$$
$${\rm{Favorable}}\,\,{\rm{ = }}\,\,{\rm{10}} \cdot {\rm{1}}\,\,{\rm{pairs}}\,{\rm{among}}\,{\rm{the}}\,{\rm{above}}\,\,\,\,\,\left[ {\,\left( {{1^{{\rm{st}}}} = {\rm{any}}\,\,{\rm{ball}}\,,\,\,{2^{{\rm{nd}}}} = {\rm{same}}\,\,{1^{{\rm{st}}}}\,\,{\rm{ball}}} \right)\,} \right]\,$$
$$? = {{10} \over {100}} = {1 \over {10}}$$

These solutions follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Portuguese-speakers :: https://www.gmath.com.br
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by regor60 » Fri Nov 09, 2018 6:30 am
BTGmoderatorDC 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

OA D

Source: Magoosh
How many ways to take a distinct pair of numbers that don't match out of 10 numbers ? 10!/2!8! = 45. This is like without replacement in order to avoid a matching pair.

How many ways to take a distinct pair out of 10 numbers with replacement ? 10^2/2 = 50. This includes the above plus matching pairs.

To find the number of matching pairs, subtract the first answer from the second: 50-45 =5

Probability of selecting a matching pair is then 5/50 = [spoiler]D,1/10[/spoiler]
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by Brent@GMATPrepNow » Sat Nov 10, 2018 8:13 am
BTGmoderatorDC 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 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,
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by Scott@TargetTestPrep » Mon Nov 12, 2018 7:18 am
BTGmoderatorDC 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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by swerve » Mon Nov 12, 2018 9:45 am
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 Ann had 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.
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