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## Coin flip questions made easy

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amitdgr Really wants to Beat The GMAT!
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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 ) 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).

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

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..

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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).

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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?

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.

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amitdgr Really wants to Beat The GMAT!
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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.

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amitdgr Really wants to Beat The GMAT!
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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?

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

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

I hope I got it right !

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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

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

I hope I got it right !
Yes, correct!

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Mon Sep 22, 2008 12:08 am
I am back with another coin-flip question.

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

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Mon Sep 22, 2008 1:14 am
amitdgr wrote:
I am back with another coin-flip question.

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.

pre-gmat Really wants to Beat The GMAT!
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Fri Sep 26, 2008 12:02 pm
WOnderful explanation Stuart:)

mberkowitz Rising GMAT Star
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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...

thanks.

amitdgr Really wants to Beat The GMAT!
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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...

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.

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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!
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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?

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Stuart Kovinsky GMAT Instructor
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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!

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pre-gmat Really wants to Beat The GMAT!
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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

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