BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 181
Joined: Sat Oct 16, 2010 1:57 pm
Thanked: 4 times

Probability

by N:Dure » Fri Dec 24, 2010 6:29 pm
1) A coin is tossed 6 times. What's the probability of getting at least 4 heads on consecutive tosses?

a) 1/64
b) 3/32
c) 15/64
d) 11/32
e) 90/64


2) Given a positive integer K

A) The probability that (K) (K+1) (K+2) (K+3) (K+4) is divisible by 20
B) 1

a) A is bigger
b) B is bigger
c) both are equal
d) Insufficient


3) Given that 'n' numbers are 8 i.e. (8,8,8,8,8....n) and 'n+2' numbers are 10. What's the arithmetic mean for both sets?
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pm
Thanked: 147 times
Followed by:3 members

by anshumishra » Fri Dec 24, 2010 6:43 pm
N:Dure wrote:1) A coin is tossed 6 times. What's the probability of getting at least 4 heads on consecutive tosses?

a) 1/64
b) 3/32
c) 15/64
d) 11/32
e) 90/64


2) Given a positive integer K

A) The probability that (K) (K+1) (K+2) (K+3) (K+4) is divisible by 20
B) 1

a) A is bigger
b) B is bigger
c) both are equal
d) Insufficient


3) Given that 'n' numbers are 8 i.e. (8,8,8,8,8....n) and 'n+2' numbers are 10. What's the arithmetic mean for both sets?

Please post one question per thread.

1). 4 consecutive heads : (1,2,3,4), (2,3,4,5) and (3,4,5,6) - 3 ways
5 consecutive heads : (1,2,3,4,5) and (2,3,4,5,6) - 2 ways
6 consecutive heads : (1,2,3,4,5,6) - 1 ways

So, required probability = 6/64 = 3/32

2). (K) (K+1) (K+2) (K+3) (K+4) is divisible by 2,3,4,5 => It is divisible by 20
So, P [(K) (K+1) (K+2) (K+3) (K+4) is divisible by 20] = 1

Hence answer is C.

3).Arithmetic mean = [8*n+10*(n+2)]/[n+(n+2)] = [18n+20]/[2n+2] = (9n+10)/(n+1).
Last edited by anshumishra on Fri Dec 24, 2010 8:02 pm, edited 1 time in total.
Thanks
Anshu

(Every mistake is a lesson learned )
Top
Master | Next Rank: 500 Posts
Posts: 181
Joined: Sat Oct 16, 2010 1:57 pm
Thanked: 4 times

by N:Dure » Fri Dec 24, 2010 7:57 pm
Namaste! Thanks Anush for your help.

Just to reiterate, in no.2 you assume K=1 so 2,3, 4, 5 making them multiples and 20 is 2*2*5 so they have common factors, and the product of Ks is also / 20, right?
I also thought the same for no.3 but the answer is (9n+10)/(n+1) don't ask me how
Last edited by N:Dure on Fri Dec 24, 2010 8:03 pm, edited 1 time in total.
Top
Master | Next Rank: 500 Posts
Posts: 181
Joined: Sat Oct 16, 2010 1:57 pm
Thanked: 4 times

by N:Dure » Fri Dec 24, 2010 8:02 pm
Another q: If 65% of the total stamps are foreign, and 35% of the stamps with last ten years (?), and 40% are the foreign stamps ten years ago (?). If the total stamps are 1200, then what number of stamps are ten years now?
Top
User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pm
Thanked: 147 times
Followed by:3 members

by anshumishra » Fri Dec 24, 2010 8:05 pm
N:Dure wrote:Namaste! Thanks Anush for your help.

Just to reiterate, in no.2 you assume K=1 so 2,3, 4, 5 making them multiples and 20 is 2*2*5 so they have common factors, and the product of Ks is also / 20, right?
I also thought the same for no.3 but the answer is (9n+10)/(9n+1) don't ask me how
For Question no. 2 : - For any value of K, the (K) (K+1) (K+2) (K+3) (K+4)...(K+n) is divisible by all the numbers from 1 to n+1.

For Question 3 : The question was not very clear, Please check the edited answer now.
Thanks
Anshu

(Every mistake is a lesson learned )
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

by Rahul@gurome » Fri Dec 24, 2010 8:11 pm
N:Dure wrote:Namaste! Thanks Anush for your help.

Just to reiterate, in no.2 you assume K=1 so 2,3, 4, 5 making them multiples and 20 is 2*2*5 so they have common factors, and the product of Ks is also / 20, right?
I also thought the same for no.3 but the answer is (9n+10)/(n+1) don't ask me how
If you take any 5 consecutive integers, one of them is always divisible by 5 and one of them is always divisible by 4.
For example 7, 8, 9, 10, 11 - 10 is divisible by 5 and 8 is divisible by 4.
So the product of 5 sonsecutive integers will always be divisible by 5*4 = 20.

In 3rd question, average is {8*n + 10*(n+2)}/{n+(n+2)} = (9n+10 )/(n+1)
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pm
Thanked: 147 times
Followed by:3 members

by anshumishra » Fri Dec 24, 2010 9:30 pm
N:Dure wrote:Another q: If 65% of the total stamps are foreign, and 35% of the stamps with last ten years (?), and 40% are the foreign stamps ten years ago (?). If the total stamps are 1200, then what number of stamps are ten years now?
N:Dure,
Please post a new question in a separate thread, otherwise it gets lost.
Again this question is very poorly worded. Either it is not from a good source or not properly written here.
Skip it.

Or lets see if someone makes something out of this problem.
Thanks
Anshu

(Every mistake is a lesson learned )
Top
Master | Next Rank: 500 Posts
Posts: 181
Joined: Sat Oct 16, 2010 1:57 pm
Thanked: 4 times

by N:Dure » Fri Dec 24, 2010 10:43 pm
Thanks Anush & Rahul

I just check q 1 and the OA is 11/32... any clue why is that?
Top
Master | Next Rank: 500 Posts
Posts: 437
Joined: Sat Nov 22, 2008 5:06 am
Location: India
Thanked: 50 times
Followed by:1 members
GMAT Score:580

by beat_gmat_09 » Fri Dec 24, 2010 11:12 pm
Hi,
The question asks for at least 4 heads.
1st we get 4 heads remaining 2 has to be Tails. This can happen - HHHHTT, number of ways heads can occur = 6!/(2!*4!) = 15
2nd we get 5 heads remaining 1 has to be tail. This can happen - HHHHHT, number of ways heads can occur = 6!/5! = 6
3rd we get all 6 heads. This can only happen in one way i.e. HHHHHH
Probability of getting heads or tails = 1/2, occurance is 6 times, thus probability of occurance either heads or tails - (1/2)^6
We need to combine the above cases, treating them as 'OR' with the probability (1/2)^6 multiplied by the number of times they can occur.

Thus required probability -
= P(4 Heads,2 tails) + P(5 heads 1 tail) + P(all 6 heads)

=[15*(1/2)^6] + [6*(1/2)^6] + [(1/2)^6]

Take out (1/2)^6 common term

=(1/2)^6[15+6+1]
= 22/64
= 11/32
HTH
Hope is the dream of a man awake
Top
User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pm
Thanked: 147 times
Followed by:3 members

by anshumishra » Sat Dec 25, 2010 6:44 am
beat_gmat_09 wrote:Hi,
The question asks for at least 4 heads.
1st we get 4 heads remaining 2 has to be Tails. This can happen - HHHHTT, number of ways heads can occur = 6!/(2!*4!) = 15
2nd we get 5 heads remaining 1 has to be tail. This can happen - HHHHHT, number of ways heads can occur = 6!/5! = 6
3rd we get all 6 heads. This can only happen in one way i.e. HHHHHH
Probability of getting heads or tails = 1/2, occurance is 6 times, thus probability of occurance either heads or tails - (1/2)^6
We need to combine the above cases, treating them as 'OR' with the probability (1/2)^6 multiplied by the number of times they can occur.

Thus required probability -
= P(4 Heads,2 tails) + P(5 heads 1 tail) + P(all 6 heads)

=[15*(1/2)^6] + [6*(1/2)^6] + [(1/2)^6]

Take out (1/2)^6 common term

=(1/2)^6[15+6+1]
= 22/64
= 11/32
HTH
The question asks for at least 4 heads on consecutive tosses.
1st we get 4 heads remaining 2 has to be Tails. This can happen - HHHHTT, number of ways heads can occur = 6!/(2!*4!) = 15
This will contain some ways such as (HTHTHH, THHHTH, etc...) which doesn't satisfy the criteria.

Correct me if I am wrong, But that is what the question meant to me. It is not asking the probability of just getting at least 4 heads (but rather at least 4 consecutive heads : HHHHTT, THHHHT,TTHHHH,HHHHHT,THHHHH,HHHHHH).
Thanks
Anshu

(Every mistake is a lesson learned )
Top
Master | Next Rank: 500 Posts
Posts: 437
Joined: Sat Nov 22, 2008 5:06 am
Location: India
Thanked: 50 times
Followed by:1 members
GMAT Score:580

by beat_gmat_09 » Sat Dec 25, 2010 7:49 pm
anshumishra wrote: The question asks for at least 4 heads on consecutive tosses.
1st we get 4 heads remaining 2 has to be Tails. This can happen - HHHHTT, number of ways heads can occur = 6!/(2!*4!) = 15
This will contain some ways such as (HTHTHH, THHHTH, etc...) which doesn't satisfy the criteria.

Correct me if I am wrong, But that is what the question meant to me. It is not asking the probability of just getting at least 4 heads (but rather at least 4 consecutive heads : HHHHTT, THHHHT,TTHHHH,HHHHHT,THHHHH,HHHHHH).
at least 4 heads on consecutive tosses
The tosses are consecutive, not the heads getting on the tosses !
Hope is the dream of a man awake
Top
User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pm
Thanked: 147 times
Followed by:3 members

by anshumishra » Sat Dec 25, 2010 7:54 pm
beat_gmat_09 wrote:
anshumishra wrote: The question asks for at least 4 heads on consecutive tosses.
1st we get 4 heads remaining 2 has to be Tails. This can happen - HHHHTT, number of ways heads can occur = 6!/(2!*4!) = 15
This will contain some ways such as (HTHTHH, THHHTH, etc...) which doesn't satisfy the criteria.

Correct me if I am wrong, But that is what the question meant to me. It is not asking the probability of just getting at least 4 heads (but rather at least 4 consecutive heads : HHHHTT, THHHHT,TTHHHH,HHHHHT,THHHHH,HHHHHH).
at least 4 heads on consecutive tosses
The tosses are consecutive, not the heads getting on the tosses !
How does toss being consecutive adds anything to the question ? What is the difference between these two questions :

A coin is tossed 6 times. What's the probability of getting at least 4 heads on consecutive tosses?
vs
A coin is tossed 6 times. What's the probability of getting at least 4 heads ?
Thanks
Anshu

(Every mistake is a lesson learned )
Top
Master | Next Rank: 500 Posts
Posts: 437
Joined: Sat Nov 22, 2008 5:06 am
Location: India
Thanked: 50 times
Followed by:1 members
GMAT Score:580

by beat_gmat_09 » Sat Dec 25, 2010 8:03 pm
anshumishra wrote: How does toss being consecutive adds anything to the question ? What is the difference between these two questions :

A coin is tossed 6 times. What's the probability of getting at least 4 heads on consecutive tosses?
vs
A coin is tossed 6 times. What's the probability of getting at least 4 heads ?
As far as my knowledge both are same.
The consecutive notion which you are talking about requires the question to state explicitly that the tosses on the coins should be heads consecutively i.e. HHHHTT, THHHHT .....
I think the problem should state something like this - ... of getting at least 4 heads consecutively on the tosses (the tosses are consecutive here)
Hope is the dream of a man awake
Top
User avatar
Legendary Member
Posts: 543
Joined: Tue Jun 15, 2010 7:01 pm
Thanked: 147 times
Followed by:3 members

by anshumishra » Sat Dec 25, 2010 8:20 pm
beat_gmat_09 wrote:
anshumishra wrote: How does toss being consecutive adds anything to the question ? What is the difference between these two questions :

A coin is tossed 6 times. What's the probability of getting at least 4 heads on consecutive tosses?
vs
A coin is tossed 6 times. What's the probability of getting at least 4 heads ?
As far as my knowledge both are same.
The consecutive notion which you are talking about requires the question to state explicitly that the tosses on the coins should be heads consecutively i.e. HHHHTT, THHHHT .....
I think the problem should state something like this - ... of getting at least 4 heads consecutively on the tosses (the tosses are consecutive here)
Good that we are on the same page, concept-wise.

It is more about how the question is worded ? I still see "getting at least 4 heads on consecutive tosses", as HHHHTT,THHHHT, etc... and not HTHTHH, HHHTHH, etc...

It would be interesting to see how others read this question :)
Thanks
Anshu

(Every mistake is a lesson learned )
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3380
Joined: Mon Mar 03, 2008 1:20 am
Thanked: 2256 times
Followed by:1535 members
GMAT Score:800

by lunarpower » Mon Dec 27, 2010 10:32 pm
anshumishra wrote:
beat_gmat_09 wrote:
anshumishra wrote: How does toss being consecutive adds anything to the question ? What is the difference between these two questions :

A coin is tossed 6 times. What's the probability of getting at least 4 heads on consecutive tosses?
vs
A coin is tossed 6 times. What's the probability of getting at least 4 heads ?
definitely not the same -- the default meaning of the former is, without a doubt, "at least 4 heads in a row without interruption". in fact, if you took a poll of native speakers of the english language (who, as far as i understand, are not common on this forum), i would bet decent money that 100% of them would interpret the statement in this way, without any hesitation.
as far as why -- it's probably just the reflex action of assigning an essential modifier to the closest logical thing that it can modify. in that case, that means the immediate assignment of the modifier "on consecutive tosses" to the phrase "at least 4 heads", giving the desired meaning.

of course, if this problem were official, it would almost certainly be phrased in a less bulky, easier-to-understand way, such as "what's the probability of getting at least 4 consecutive heads?"

--

as another example, let's say that you enter a hallway that has a door on the right, and then a door on the left, and then a door on the right, and then a door on the left, etc.
if someone says "go to the third door on the right", you will not go to the third door overall -- you'd count 1, 2, 3 on the right, and you'd ultimately arrive at the fifth door in the hallway.
if that makes sense to you, then the above explanation should make sense as well. (and even if it doesn't, you should still commit that general form of description to memory -- since that's the way those modifiers are going to work, if you see them.)
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Top
Post new topic Post Reply