Welcome! Check out our free B-School Guides to learn how you compare with other applicants.
Login or Register
 

Coin flip questions made easy

This topic has 19 expert replies and 40 member replies
Goto page Previous Next
Bookmark
Previous Topic Next Topic
amitdgr Really wants to Beat The GMAT!
Joined
29 Jun 2008
Posted:
228 messages
Thanked:
12 times
Post Fri Sep 19, 2008 10:16 am
Stuart Kovinsky wrote:
If we want fewer (as a sentence correction aside, we use "fewer" when the noun is countable - less vs fewer is commonly tested on the GMAT Very Happy ) than 2 boys, we want 0 boys or 1 boy.

So, we go to the n=10 row (which is the 11th row down - remember that the apex of the triangle is the n=0 row) and add up the first 2 entries. The sum of the row will be 2^10.

It would actually be much quicker to do this using combinations, since we're looking at k values of 0 and 1:

10C0 = 1 (any nC0 = 1)
10C1 = 10 (any nC1 = n)

In fact, by the above principles (or just by looking at the triangle), we can see that the first two entries in every row are 1 and n (and, since the rows are symmetrical, the last two entries in every row are n and 1).

So, the answer would be:

1 - (1 + 10)/2^10

1 - 11/2^10

1024/1024 - 11/1024

(1024 - 11)/1024 = 1013/1024
Whoops !! LESS VS FEWER !! Thanks for noting my mistake and actually letting me know Smile

so to find cases where we have FEWER than 2 boys , we actually follow you initial formula of nCk/2^n

so for 0 boy(s) = 10C0/2^10
for 1 boy = 10C1/2^10

1- [10C0/2^10 + 10C1/2^10] = 1013/1024 ...

Great ...Things getting pretty clear now... Thanks a lot ! I think I like the formula approach better ...

Is there a possibility of GMAT asking probability questions such as

There are 30% chances of boy being born and 70% chances of girl being born...Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?

That would definitely complicate things a bit, Cuz we cant use the above discussed methods then.....

I hope such problems are not what GMAT is out to test us upon...

Thanks for being such a great help Stuart..

_________________
Please visit my blog Gmat Tips and Strategies

Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!

GMAT/MBA Expert

Stuart Kovinsky GMAT Instructor
Joined
08 Jan 2008
Posted:
3156 messages
Followed by:
460 members
Thanked:
1431 times
GMAT Score:
800
Post Fri Sep 19, 2008 11:47 am
amitdgr wrote:
Is there a possibility of GMAT asking probability questions such as

There are 30% chances of boy being born and 70% chances of girl being born...Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?

That would definitely complicate things a bit, Cuz we cant use the above discussed methods then.....

I hope such problems are not what GMAT is out to test us upon...

Thanks for being such a great help Stuart..
Even your original question (with the 50/50 chance) is likely more difficult than anything you'll see on the GMAT. This version is WAY harder than any GMAT probability question (mostly because it requires calculations that are too time consuming to do by hand).

_________________
Stuart Kovinsky, B.A. LL.B.
Kaplan Test Prep & Admissions
Toronto Office
1-800-KAP-TEST
www.kaptest.com

Learn more about me

Free GMAT Practice Test under Proctored Conditions! - Find a practice test near you or live and online in Kaplan's Classroom Anywhere environment. Register today!

GMAT/MBA Expert

Ian Stewart GMAT Instructor
Joined
02 Jun 2008
Posted:
2178 messages
Followed by:
245 members
Thanked:
950 times
GMAT Score:
780
Post Fri Sep 19, 2008 12:46 pm
amitdgr wrote:
Is there a possibility of GMAT asking probability questions such as

There are 30% chances of boy being born and 70% chances of girl being born...Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?


I'm about 75% sure that I have seen a GMAT question like this. If I recall correctly, I saw a question testing this idea in GMATFocus, but I see a lot of GMAT-style questions from various sources, so I'm not completely sure that was the source (I might instead be remembering something I saw on a forum or from another place). As Stuart points out, almost any question like this would require an awkward amount of calculation to complete, but the question I recall seeing had answers that were not simplified, so it was easy to answer in under two minutes- as soon as you saw how to set up the calculation, you were done. The numbers were a bit easier than those above- the question was something more like:

In Country X, each time a family has a child, there is a 30% chance that the child is a boy and a 70% chance the child is a girl. If a family in Country X will have five children, what is the probability exactly one of the children will be a girl?

with answer choices like:

0.7
(0.3)^4
(0.7)*(0.3)^4
etc.

None of those answers is right, but I'll leave the question unsolved for now in case anyone wants to give it a shot.

_________________
Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

private GMAT tutor in Toronto

amitdgr Really wants to Beat The GMAT!
Joined
29 Jun 2008
Posted:
228 messages
Thanked:
12 times
Post Fri Sep 19, 2008 7:56 pm
Stuart Kovinsky wrote:
Even your original question (with the 50/50 chance) is likely more difficult than anything you'll see on the GMAT. This version is WAY harder than any GMAT probability question (mostly because it requires calculations that are too time consuming to do by hand).
Thanks Stuart. I hope you are right. It would not be very exciting to see way too complicated questions on the exam.

_________________
Please visit my blog Gmat Tips and Strategies

amitdgr Really wants to Beat The GMAT!
Joined
29 Jun 2008
Posted:
228 messages
Thanked:
12 times
Post Fri Sep 19, 2008 8:01 pm
Ian Stewart wrote:
I'm about 75% sure that I have seen a GMAT question like this. If I recall correctly, I saw a question testing this idea in GMATFocus, but I see a lot of GMAT-style questions from various sources, so I'm not completely sure that was the source (I might instead be remembering something I saw on a forum or from another place). As Stuart points out, almost any question like this would require an awkward amount of calculation to complete, but the question I recall seeing had answers that were not simplified, so it was easy to answer in under two minutes- as soon as you saw how to set up the calculation, you were done. The numbers were a bit easier than those above- the question was something more like:

In Country X, each time a family has a child, there is a 30% chance that the child is a boy and a 70% chance the child is a girl. If a family in Country X will have five children, what is the probability exactly one of the children will be a girl?

with answer choices like:

0.7
(0.3)^4
(0.7)*(0.3)^4
etc.

None of those answers is right, but I'll leave the question unsolved for now in case anyone wants to give it a shot.
I managed to dig out the approach from your post here http://www.beatthegmat.com/probability-of-rain-t14715.html

Smile

5C1 * (0.7)^1 * (0.3)^4

I hope I got it right !

_________________
Please visit my blog Gmat Tips and Strategies

GMAT/MBA Expert

Ian Stewart GMAT Instructor
Joined
02 Jun 2008
Posted:
2178 messages
Followed by:
245 members
Thanked:
950 times
GMAT Score:
780
Post Sat Sep 20, 2008 6:16 am
amitdgr wrote:
I managed to dig out the approach from your post here http://www.beatthegmat.com/probability-of-rain-t14715.html

Smile

5C1 * (0.7)^1 * (0.3)^4

I hope I got it right !
Yes, correct! Smile

amitdgr Really wants to Beat The GMAT!
Joined
29 Jun 2008
Posted:
228 messages
Thanked:
12 times
Post Mon Sep 22, 2008 12:08 am
I am back with another coin-flip question. Mr. Green

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.

a) 5/16
b) 15/32
c) 1/2
d) 21/32
e) 11/16

_________________
Please visit my blog Gmat Tips and Strategies

Morgoth GMAT Destroyer!
Joined
22 Sep 2008
Posted:
316 messages
Thanked:
34 times
Target GMAT Score:
750+
Post Mon Sep 22, 2008 1:14 am
amitdgr wrote:
I am back with another coin-flip question. Mr. Green

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.

a) 5/16
b) 15/32
c) 1/2
d) 21/32
e) 11/16
IMO 15/32

"What is the probability that Kate has more than $10 but less than $15"

There can be only 2 cases

HTTTT [kate looses $1 and gains $4 = $13] = 5/32 [exactly 4 tails, 1,5,10,10,5,1]

HHTTT [kate looses $2 and gains $3 = $11] = 10/32 [exactly, 3 tails 1,5,10,10,5,1]


5/32 + 10/32 = 15/32

Hence B.

Whats the answer.

pre-gmat Really wants to Beat The GMAT! Default Avatar
Joined
15 Jul 2008
Posted:
105 messages
Thanked:
4 times
Post Fri Sep 26, 2008 12:02 pm
WOnderful explanation Stuart:)

mberkowitz Rising GMAT Star Default Avatar
Joined
20 Jul 2008
Posted:
76 messages
Thanked:
1 times
Post Fri Sep 26, 2008 8:31 pm
"1 - (1 + 10)/2^10

1 - 11/2^10

1024/1024 - 11/1024

(1024 - 11)/1024 = 1013/1024"

can you explain this last step, specifically why it is 1 minus ... i would expect it to be 11/2^10 rather than 1- 11...

very helpful post thus far...

thanks.

amitdgr Really wants to Beat The GMAT!
Joined
29 Jun 2008
Posted:
228 messages
Thanked:
12 times
Post Fri Sep 26, 2008 10:25 pm
mberkowitz wrote:
"1 - (1 + 10)/2^10

1 - 11/2^10

1024/1024 - 11/1024

(1024 - 11)/1024 = 1013/1024"

can you explain this last step, specifically why it is 1 minus ... i would expect it to be 11/2^10 rather than 1- 11...

very helpful post thus far...

thanks.
Ok i'll try to explain.

The question : The probability of a new born baby is a boy is 50% and that of a new born baby is a girl is 50%. Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?

ten babies are going to be born. we want cases where at least 2 boys are born, at least 2 means 2 or more

so we have to add up the following cases to find the probability of at least 2 boys being born.

1) 2 boys and 8 girls
2) 3 boys and 7 girls
3) 4 boys and 6 girls
4) 5 boys and 5 girls
5) 6 boys and 4 girls
6) 7 boys and 3 girls
7) 8 boys and 2 girls
8 ) 9 boys and 1 girl
9) 10 boys and 0 girl

Pretty tedious task ain't it ??

Luckily probability has this small but VERY useful rule.

The sum of all possibilities is 1.

We can use this powerful rule to simplify our problem above.

so the possibility of at least 2(in other words 2 or more) boys being born + possibility of FEWER than 2 boys being born = 1

possibility of at least 2(in other words 2 or more) boys being born = 1 - possibility of FEWER than 2 boys being born

Now possibility of FEWER than 2 boys being born is the sum of

a) 0 boys and 10 girls (10C0/2^10)
b) 1 boy and 9 girls (0C1/2^10 )

so possibility of at least 2(in other words 2 or more) boys being born = 1- [10C0/2^10 + 10C1/2^10] = 1013/1024

Hope this helps.

_________________
Please visit my blog Gmat Tips and Strategies

mberkowitz Rising GMAT Star Default Avatar
Joined
20 Jul 2008
Posted:
76 messages
Thanked:
1 times
Post Sat Sep 27, 2008 5:27 am
thanks, i was looking at the questions p(0) or p(1) rather than p(2-10). appreciate the help.

pre-gmat Really wants to Beat The GMAT! Default Avatar
Joined
15 Jul 2008
Posted:
105 messages
Thanked:
4 times
Post Mon Oct 13, 2008 9:25 pm
Stuart,

Is the above strategy applicable to only scenarios where the probability of Heads and tails is 50% each? One of the questions i got on GMAT Practice test GMat prep 2 ( i know the answer is posted elsewhere on the forum but wanted to ask you about it) is following:

For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads 'at least' 4 times?

(A) (0.6)^5

(B) 2(0.6)^4

(C) 3[(0.6)^4](0.4)

(D) 4[((0.6)^4)(0.4)] + (0.6)^5

(E) 5[((0.6)^4)(0.4)] + (0.6)^5

When I saw the coin flip question, i immediately tried using the pascals traingle. It did not work. is there a way to use the pascals strategy on this one?

GMAT/MBA Expert

Stuart Kovinsky GMAT Instructor
Joined
08 Jan 2008
Posted:
3156 messages
Followed by:
460 members
Thanked:
1431 times
GMAT Score:
800
Post Tue Oct 14, 2008 11:55 am
pre-gmat wrote:
Stuart,

Is the above strategy applicable to only scenarios where the probability of Heads and tails is 50% each? One of the questions i got on GMAT Practice test GMat prep 2 ( i know the answer is posted elsewhere on the forum but wanted to ask you about it) is following:

For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads 'at least' 4 times?

(A) (0.6)^5

(B) 2(0.6)^4

(C) 3[(0.6)^4](0.4)

(D) 4[((0.6)^4)(0.4)] + (0.6)^5

(E) 5[((0.6)^4)(0.4)] + (0.6)^5

When I saw the coin flip question, i immediately tried using the pascals traingle. It did not work. is there a way to use the pascals strategy on this one?
We can still use the triangle in part, but the formula will be different.

We have 5 flips and want at least 4 heads, i.e. 4 or 5 heads.

If we didn't care about the order, the chance of getting 4H and 1T would be:

.6 * .6 * .6 * .6 * .4 = (.6)^4 * .4

However, we do care about the order. Looking at the n=5 row, we see that the 2nd last entry is 5 - so there are 5 different ways to get exactly 4 heads out of 5 flips.

So, the chance of getting exactly 4H is 5 * ((.6)^4 * .4)

Now we have to add the chance of getting exactly 5H, which is simply (.6)^5.

So, our final answer is 5 * ((.6)^4 * .4) + (.6)^5... choose (E).

Note that strategic elimination will get us to the correct answer in about 30 seconds:

We know that we want 4H or 5H, so there will be addition involved... eliminate (a), (b) and (c).

Looking at the triangle (or just by applying the combinations formula or common sense), we see that 5C4 = 5, so the multiplier for the first part of the expression will be "5"... eliminate (d).

Only (e) remains, pick it confidently and do the happy dance!

_________________
Stuart Kovinsky, B.A. LL.B.
Kaplan Test Prep & Admissions
Toronto Office
1-800-KAP-TEST
www.kaptest.com

Learn more about me

Free GMAT Practice Test under Proctored Conditions! - Find a practice test near you or live and online in Kaplan's Classroom Anywhere environment. Register today!
pre-gmat Really wants to Beat The GMAT! Default Avatar
Joined
15 Jul 2008
Posted:
105 messages
Thanked:
4 times
Post Wed Oct 15, 2008 3:48 pm
Thanks Stuart,

QUick Question: WHy do we care about Orders here? Is it because we are following Pascals Triangle methodology?



Last edited by pre-gmat on Wed Oct 15, 2008 3:57 pm; edited 1 time in total

Goto page Previous Next

Best Conversation Starters
1 varun289 43 topics
2 greenwich 30 topics
3 sana.noor 22 topics
4 guerrero 20 topics
5 killerdrummer 18 topics
See More Top Beat The GMAT Members...

Most Active Experts
1 image description Brent@GMATPrepNow

GMAT Prep Now Teacher

 205 posts
2 image description Anju@Gurome

Gurome

 146 posts
3 image description GMATGuruNY

The Princeton Review Teacher

 145 posts
4 image description Jim@StratusPrep

Stratus Prep

 96 posts
5 image description David@VeritasPrep

Veritas Prep

 39 posts
See More Top Beat The GMAT Experts