A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digits 6?
A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000
OA:B
What is the best approach to solve the question?
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Hard probability question
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Jay@ManhattanReview
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Hi Mo2men,Mo2men wrote:A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digits 6?
A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000
OA:B
What is the best approach to solve the question?
There are 5 digits and we can use 0-9 digits any number of times. So, if there is no condition, there are 10*10*10*10*10 = 10^5 = 100,000 passwords possible. This is our denominator for the probability.
We want three 6 out of 5 digits.
Ways of choosing three 6s and two other and 6s = 1*1*1*9*9 = 81.
Let's arrange the three 6s. They can be arranged in 5C3 = 5C2 = (5*4)/(1*2) = 10 ways.
Total number of ways = 81*10 = 810. This is our numerator for the probability.
Thus, the required probability = [spoiler]810 / 100000.[/spoiler]
The correct answer: B
Hope this helps!
Relevant book: Manhattan Review GMAT Combinatorics and Probability Guide
-Jay
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Mo2men
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Hi Jay,Jay@ManhattanReview wrote: Hi Mo2men,
Ways of choosing three 6s and two other and 6s = 1*1*1*9*9 = 81.
Let's arrange the three 6s. They can be arranged in 5C3 = 5C2 = (5*4)/(1*2) = 10 ways.
Total number of ways = 81*10 = 810. This is our numerator for the probability.
Thanks for your help.
As mentioned above, you calculated the way and then arrangement, could we solve it using permutation? Permutation cares about ways and places.
Thanks
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Jay@ManhattanReview
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Hi Mo2men,Mo2men wrote:Hi Jay,Jay@ManhattanReview wrote: Hi Mo2men,
Ways of choosing three 6s and two other and 6s = 1*1*1*9*9 = 81.
Let's arrange the three 6s. They can be arranged in 5C3 = 5C2 = (5*4)/(1*2) = 10 ways.
Total number of ways = 81*10 = 810. This is our numerator for the probability.
Thanks for your help.
As mentioned above, you calculated the way and then arrangement, could we solve it using permutation? Permutation cares about ways and places.
Thanks
Yes, we can do it, but that's a labored approach and not advisable.
For your interest, we can do this way.
The number of ways three 6s and two other than 6 digits can be arranged in 5P5 / 3! ways = 5! / 3! = 20 ways.
We know that if there are a few indistinguishable objects, we must divide the total number of ways by the factorial value of their repetition.
Here, there are three 6s, so we divided 5P5 by 3!.
However, 5! / 3! = 20 ways is still not what we desire. The 20 ways would include those passwords that have two 0s, two 1s, two 2s, two 3s, two 4s, two 5s, No 6s, two 7s, two 8s, two 9s, when swapped at their fixed positions, though it does not give us new arrangement, give new ways; so we must get rid of these. Remember that a password such as 66600 is wanted, but a password such as 66600 is to be excluded as it is the same as the former. That's the reason I applied combination in my solution. The combination does the selection and takes care of repetitions.
Number of dummy ways = 5! / (3!*2!) = 10 ways. I divided 5! / 3! by '2!' to find out the repetitions.
So, the number of ways three 6s and two other than 6 digits can be arranged in = 20 - 10 = 10 ways.
Thus, total number of such passwords = 81*10 = 810. It's not advisable to do this way. It's unduly complicated.
Hope this makes sense.
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Brent@GMATPrepNow
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We can also solve this question by applying some probability rules along with a counting technique.Mo2men wrote:A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digits 6?
A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000
OA:B
Let's start by finding a very specific probability: P(1st digit is 6 AND 2nd digit is 6 AND 3rd digit is 6 AND 4th digit is not 6 AND 5th digit is not 6)
This satisfies the condition that there are exactly 3 sixes.
P(1st digit is 6 AND 2nd digit is 6 AND 3rd digit is 6 AND 4th digit is not 6 AND 5th digit is not 6) = P(1st digit is 6) x P(2nd digit is 6) x P(3rd digit is 6) x P(4th digit is not 6) x P(5th digit is not 6)
= 1/10 x 1/10 x 1/10 x 9/10 x 9/10
= 81/100,000
Of course, this is just the probability for P(6 - 6 - 6 - not6 - not6)
Notice that other configurations will have the same probability.
For example, P(6 - 6 - not6 - not6 - 6) = 81/100,000
For example, P(not6 - 6 - 6 - not6 - 6) = 81/100,000
And so on.
In how many ways can we arrange three 6's and two non-6's?
One way to answer this question is to ask "In how many ways can I select 3 places to hold the 6's?"
Once we select the 3 places to hold the 6's, the 2 remaining places will hold the non-6's.
Well, we can select 3 places from 5 places in 5C3 ways
5C3 = 10
So, there are 10 different ways in which we can satisfy the condition that we have exactly three 6's.
So, P(exactly three 6's) = (10)(81/100,000)
= 810/100,000
Answer: B
Cheers,
Brent
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Hi Mo2men,
If you're not sure how to approach a given question, then it's sometimes best to do a little work and look for a pattern (since there's almost always a pattern involved in each GMAT question).
To start, the total number of 5-digit codes is (10)^5 = 100,000 options. We're asked to consider 5-digit codes that have EXACTLY three 6s in them, so let's talk about how that might occur...
6 6 6 (not6) (not6) = there are (1)(1)(1)(9)(9) = 81 codes that fit this pattern
What if I put the three 6s in slightly different 'spots' though...
6 6 (not6) 6 (not6) = there are (1)(1)(9)(1)(9) = 81 codes that fit that pattern TOO.
It stands to reason that each "three 6s option" will give us 81 codes, so we're going to end up with a TOTAL number of codes that is a multiple of 81. Since the answer choices have clearly NOT been reduced, the correct answer stands out...
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
If you're not sure how to approach a given question, then it's sometimes best to do a little work and look for a pattern (since there's almost always a pattern involved in each GMAT question).
To start, the total number of 5-digit codes is (10)^5 = 100,000 options. We're asked to consider 5-digit codes that have EXACTLY three 6s in them, so let's talk about how that might occur...
6 6 6 (not6) (not6) = there are (1)(1)(1)(9)(9) = 81 codes that fit this pattern
What if I put the three 6s in slightly different 'spots' though...
6 6 (not6) 6 (not6) = there are (1)(1)(9)(1)(9) = 81 codes that fit that pattern TOO.
It stands to reason that each "three 6s option" will give us 81 codes, so we're going to end up with a TOTAL number of codes that is a multiple of 81. Since the answer choices have clearly NOT been reduced, the correct answer stands out...
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
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Mo2men
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Dear Brent,Brent@GMATPrepNow wrote:
We can also solve this question by applying some probability rules along with a counting technique.
Let's start by finding a very specific probability: P(1st digit is 6 AND 2nd digit is 6 AND 3rd digit is 6 AND 4th digit is not 6 AND 5th digit is not 6)
This satisfies the condition that there are exactly 3 sixes.
P(1st digit is 6 AND 2nd digit is 6 AND 3rd digit is 6 AND 4th digit is not 6 AND 5th digit is not 6) = P(1st digit is 6) x P(2nd digit is 6) x P(3rd digit is 6) x P(4th digit is not 6) x P(5th digit is not 6)
= 1/10 x 1/10 x 1/10 x 9/10 x 9/10
= 81/100,000
Of course, this is just the probability for P(6 - 6 - 6 - not6 - not6)
Notice that other configurations will have the same probability.
For example, P(6 - 6 - not6 - not6 - 6) = 81/100,000
For example, P(not6 - 6 - 6 - not6 - 6) = 81/100,000
And so on.
In how many ways can we arrange three 6's and two non-6's?
One way to answer this question is to ask "In how many ways can I select 3 places to hold the 6's?"
Once we select the 3 places to hold the 6's, the 2 remaining places will hold the non-6's.
Well, we can select 3 places from 5 places in 5C3 ways
5C3 = 10
So, there are 10 different ways in which we can satisfy the condition that we have exactly three 6's.
So, P(exactly three 6's) = (10)(81/100,000)
= 810/100,000
Answer: B
Cheers,
Brent
Thanks for your help.I solved the question almost as you did until the step of 5C3.
what really confused me is why we use 5C3 not 5P3? what is behind use combination here? I feel that permutation is used because we care about places of 6's in that question.
Can you elaborate more about using combination not permutation ?
Thanks
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P(exactly n times) = P(one way) * total possible ways.Mo2men wrote:A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digits 6?
A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000
OA:B
What is the best approach to solve the question?
Let S = six and N = not six.
Since there are 10 digits, P(six) = 1/10 and P(not six) = 9/10.
P(one way):
One way to get an equal three sixes:
SSSNN.
P(1st digit is S) = 1/10.
P(2nd digit is S) = 1/10.
P(3rd digit is S) = 1/10.
P(4th digit is N) = 9/10.
P(5th digit is N) = 9/10.
Since we want all of these events to happen, we MULTIPLY:
1/10 * 1/10 * 1/10 * 9/10 * 9/10 = 81/100000.
Total possible ways:
SSSNN is only ONE WAY to get exactly three sixes
Now we must account for ALL OF THE WAYS to get exactly three sixes.
Any arrangement of the letters SSSNN represents one way to get exactly three sixes.
Thus, to account for ALL OF THE WAYS to get exactly three sixes, the result above must be multiplied by the number of ways to arrange the letters SSSNN.
Number of ways to arrange 5 elements = 5!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 3! to account for the three identical S's and by 2! to account for the two identical N's:
5!/(3!2!) = 10.
Multiplying the results above, we get:
P(exactly three sixes) = 81/100000 * 10 = 810/100000.
The correct answer is B.
More practice:
https://www.beatthegmat.com/select-exact ... 88786.html
https://www.beatthegmat.com/probability- ... 14250.html
https://www.beatthegmat.com/a-single-par ... 28342.html
https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html
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Brent@GMATPrepNow
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That's a good (and often asked) question!Mo2men wrote: Dear Brent,
Thanks for your help.I solved the question almost as you did until the step of 5C3.
what really confused me is why we use 5C3 not 5P3? what is behind use combination here? I feel that permutation is used because we care about places of 6's in that question.
Can you elaborate more about using combination not permutation ?
Thanks
The usual response is to ask whether order matters, but I believe this often complicates matters further.
To understand what I mean, consider this question:
There are 7 students. In how many ways can we elect a President and Treasurer, if no student can hold both positions?
Does the order of the selected students matter?
Hard to say. Does it matter if we select the Treasurer first and the President second?
No.
So, we use combinations (7C2) then, right?
No.
A few years ago, I took a stab at a different way of determining whether or not to use combinations.
Here's the article: https://www.gmatprepnow.com/articles/do ... 3-part-iii
Cheers,
Brent
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Matt@VeritasPrep
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Not to speak for Brent, but the idea here is that you're CHOOSING the three places that will be 6. Once you've chosen them, they can't be ordered, so you're done.Mo2men wrote: Dear Brent,
Thanks for your help.I solved the question almost as you did until the step of 5C3.
what really confused me is why we use 5C3 not 5P3? what is behind use combination here? I feel that permutation is used because we care about places of 6's in that question.
Can you elaborate more about using combination not permutation ?
Thanks
That's the crucial distinction between a permutation and a combination. In a combination, you merely choose; in a permutation, you choose, THEN arrange your choices.
