Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?
5/16
15/32
½
21/32
11/16
probability
This topic has expert replies
- anshumishra
- Legendary Member
- Posts: 543
- Joined: Tue Jun 15, 2010 7:01 pm
- Thanked: 147 times
- Followed by:3 members
Kate will have more than 10$ and less than 15$, if she wins 3 or 4 times.rtaha2412 wrote:Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?
5/16
15/32
½
21/32
11/16
So, probability of getting 3 tails or 4 tails = 5C3/2^5 + 5C4/2^5 = 10/32 + 5/32 = 15/32.
B
Thanks
Anshu
(Every mistake is a lesson learned )
Anshu
(Every mistake is a lesson learned )
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.rtaha2412 wrote:Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?
5/16
15/32
½
21/32
11/16
When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2
P(5 tails) = 1/32
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.
The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
-
- Master | Next Rank: 500 Posts
- Posts: 298
- Joined: Tue Feb 16, 2010 1:09 am
- Thanked: 2 times
- Followed by:1 members
Kate already has $10 . So for the probability that at the end he must have more that $ 10 but less than $ 15 , at least one but not 5 tails should have appeared . So , why aren't we considering 1 and 2 appearances that would in turn earn him 1 or 2 dollars.GMATGuruNY wrote:As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.rtaha2412 wrote:Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?
5/16
15/32
½
21/32
11/16
When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2
P(5 tails) = 1/32
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.
The correct answer is B.
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
For every tails, Kate gets $1; for every heads, she loses $1.Deepthi Subbu wrote:Kate already has $10 . So for the probability that at the end he must have more that $ 10 but less than $ 15 , at least one but not 5 tails should have appeared . So , why aren't we considering 1 and 2 appearances that would in turn earn him 1 or 2 dollars.GMATGuruNY wrote:As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.rtaha2412 wrote:Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?
5/16
15/32
½
21/32
11/16
When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2
P(5 tails) = 1/32
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.
The correct answer is B.
0 tails, 5 heads: Kate has 10+0-5 = $5
1 tails, 4 heads: Kate has 10+1-4 = $7
2 tails, 3 heads: Kate has 10+2-3 = $9
3 tails, 2 heads: Kate has 10+3-2 = $11
4 tails, 1 heads: Kate has 10+4-1 = $13
5 tails, 0 heads: Kate has 10+5-0 = $15
Only 3 tails and 4 tails will give Kate between $10 and $15.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
-
- Master | Next Rank: 500 Posts
- Posts: 298
- Joined: Tue Feb 16, 2010 1:09 am
- Thanked: 2 times
- Followed by:1 members
Oh ya , got itGMATGuruNY wrote:For every tails, Kate gets $1; for every heads, she loses $1.Deepthi Subbu wrote:Kate already has $10 . So for the probability that at the end he must have more that $ 10 but less than $ 15 , at least one but not 5 tails should have appeared . So , why aren't we considering 1 and 2 appearances that would in turn earn him 1 or 2 dollars.GMATGuruNY wrote:As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.rtaha2412 wrote:Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?
5/16
15/32
½
21/32
11/16
When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2
P(5 tails) = 1/32
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.
The correct answer is B.
0 tails, 5 heads: Kate has 10+0-5 = $5
1 tails, 4 heads: Kate has 10+1-4 = $7
2 tails, 3 heads: Kate has 10+2-3 = $9
3 tails, 2 heads: Kate has 10+3-2 = $11
4 tails, 1 heads: Kate has 10+4-1 = $13
5 tails, 0 heads: Kate has 10+5-0 = $15
Only 3 tails and 4 tails will give Kate between $10 and $15.
-
- Legendary Member
- Posts: 712
- Joined: Fri Sep 25, 2015 4:39 am
- Thanked: 14 times
- Followed by:5 members
Dear GMATGuruGMATGuruNY wrote:As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.rtaha2412 wrote:Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?
5/16
15/32
½
21/32
11/16
When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2
P(5 tails) = 1/32
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.
The correct answer is B.
I encountered another solution in which total possibility = 2^5 (which will include a certain number of similar outcomes in different orders)
Since it does not matter which of the 5 tosses is won, as long as it's either 3 or 4 of them, we can use the combination formula:
For 3 wins, we have 5!/3!2! = 10 possible combinations of 3 wins
For 4 wins, we have 5!/4!1! = 5 possible combinations of 4 wins
My question here:
How we can divide combinations over permutation? as As far as I know, we must either deal with all combinations (in numerator and denominator) or all permutations (in numerator and denominator).
Should not we discard similar outcome from 2^5 to make numerator matches the denominator? or the other way to consider the order in the numerator to match the order in the denominator?
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
3 wins and 2 losses = WWWLL.Mo2men wrote:[Dear GMATGuru
I encountered another solution in which total possibility = 2^5 (which will include a certain number of similar outcomes in different orders)
Since it does not matter which of the 5 tosses is won, as long as it's either 3 or 4 of them, we can use the combination formula:
For 3 wins, we have 5!/3!2! = 10 possible combinations of 3 wins
For 4 wins, we have 5!/4!1! = 5 possible combinations of 4 wins
My question here:
How we can divide combinations over permutation? as As far as I know, we must either deal with all combinations (in numerator and denominator) or all permutations (in numerator and denominator).
One way to determine the number of ways to get 3 wins is to count the number of ways to arrange the 5 letters WWWLL.
The 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 3 identical W's and by 2! to account for the 2 identical L's:
5!/(3!2!) = 10
Another approach:
There are 5 positions in the arrangement.
Each position must be occupied by W or L.
A good arrangement occurs when W's occupy A COMBINATION OF 3 POSITIONS in the 5-position arrangement.
Thus, to determine the number of ways to get 3 wins, we can count the number ways to choose a combination of 3 positions from 5 options:
5C3 = 5!/(3!2!) = 10
Each approach yields the same result.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3