pre-gmat wrote:Thanks Stuart,

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

We care about order because it shows us the different ways we can get 4 out of 5 heads, which we need to take into account in our calculation. You don't need to think of it that way (we can also just think about how many different subgroups of 4 we can make out of 5 total entities).

Thanks:)

(question from someone terrible in math)

Does it make sense to apply this formula on below question from OG11? If so, how can I do it? I have tried with no success.

From OG11 page 241 ex 173

THE PROBABILITY IS 1/2 THAT A CERTAIN COIN WILL TURN UP HEADS ON ANY GIVEN TOSS. IF THE COIN IS TO BE TOSSED THREE TIMES, WHAT IS THE PROBABILITY THAT ON AT LEAST ONE OF THE TOSSES THE COIN WILL TURN UP TAILS?

A) 1/8
B) 1/2
C) 3/4
D) 7/8
E) 15/16

GMATCHPOINT wrote:(question from someone terrible in math)

Does it make sense to apply this formula on below question from OG11? If so, how can I do it? I have tried with no success.

From OG11 page 241 ex 173

THE PROBABILITY IS 1/2 THAT A CERTAIN COIN WILL TURN UP HEADS ON ANY GIVEN TOSS. IF THE COIN IS TO BE TOSSED THREE TIMES, WHAT IS THE PROBABILITY THAT ON AT LEAST ONE OF THE TOSSES THE COIN WILL TURN UP TAILS?

A) 1/8
B) 1/2
C) 3/4
D) 7/8
E) 15/16

The formula works very well for this question. First, let's simplify.

If we want AT LEAST 1 tails, then the only thing that we DO NOT want is all 3 heads.

Remember this general formula:

Prob (want) = 1 - Prob (don't want)

Applying that to this question:

Prob (at least 1 tails) = 1 - Prob (all 3 heads)

Prob(all 3 heads) = 3C3/2^3 = 1/8

Prob (at least 1 tails) = 1 - 1/8 = 7/8

Of course, you could also answer this question without the forumla, using a bit of logic and common sense.

The only result we don't want is HHH. For 3 flips, there are 2^3 = 8 total possible outcomes. If there's 1 outcome we don't want, there must be 7 outcomes that we do want, so the chance of getting what we want is 7/8.

You could also look at the n=3 row of Pascal's Triangle. The only thing we don't want is 0 tails, so if we add up 1 tails + 2 tails + 3 tails we get 3 + 3 + 1 = 7. The whole row totals to 8, so again we end up with an answer of 7/8.

I think the strategy outlined is great. However, for coin problems or any problems where you have multiple trials and the probabilities stay the same I like to use bernouli's formula:
(nCr)p^r x q^n-r

n = number of trials
r = number of specific events you wish to obtain
p = probability that the event will occur
q = probability that the event will not occur

I know it looks scary but it's easy to use.
Here are the steps to using it on this problem.

Question:

THE PROBABILITY IS 1/2 THAT A CERTAIN COIN WILL TURN UP HEADS ON ANY GIVEN TOSS. IF THE COIN IS TO BE TOSSED THREE TIMES, WHAT IS THE PROBABILITY THAT ON AT LEAST ONE OF THE TOSSES THE COIN WILL TURN UP TAILS?

A) 1/8
B) 1/2
C) 3/4
D) 7/8
E) 15/16

Solution:
At least one tail means 1 tail or 2 tails or 3tails. OR means we have to add probabilities.
P(1 tail) + P(2tails) + P(3 Tails)
We can solve this but have to find 3 probabilities. Not fun.


When you see the word at least immediately think of the reverse situation as it saves time.
In reverse what's the condition where we don't have at least 1 tail. That would be HHH or all heads.
We find the probability of the event not occurring( i.e all heads) and minus this from 1.

P(at least one tail) = 1 - (event not occuring)
To use bernouli we write down:
sucesses: P(H) = 1/2 We want 3 heads
failures: P(T) = 1/2 We want 0 tails
number of trials = 3 flips
We want 3 heads and 0 tails
Our successes in this case would be getting a head. So out of 3 flips we want to choose 3 heads or 3C3.
Here's what the formula looks like:
(1/2)³ x (1/2)^0 x 3C3
(1/2)³ x (1/2)^0 = 1/8
3C3 = 3!/3!0!= 1
So the probability of getting 3 heads and zero tails after 3 trials is 1/8
Therefore,
P(at least one tail) = 1 - 1/8 = 7/8

nice. many tks!

Stuart Kovinsky wrote:
Let's look at a much more complicated question:

"A fair coin is flipped 5 times. What's the probability of getting at least 2 heads?"

If we want at least 2 heads, we want 2 heads, 3 heads, 4 heads OR 5 heads. Pretty much whenever we see "OR" in probability, we add the individual probabilities.

Looking at the 5 flip row, we have 1 5 10 10 5 1. For 2H, 3H, 4H and 5H we add up the 3rd, 4th, 5th and 6th entries: 10+10+5+1=26.

Summing the whole row, we get 32.

So, the chance of getting at least 2 heads out of 5 flips is 26/32

Using nCk/2^n

for one HEADS = 5C1/2^5 = 5/32

So for at least two:

1-(5/32) = 27/32

What am I missing here ?

Let's look at a much more complicated question:

"A fair coin is flipped 5 times. What's the probability of getting at least 2 heads?"

If we want at least 2 heads, we want 2 heads, 3 heads, 4 heads OR 5 heads. Pretty much whenever we see "OR" in probability, we add the individual probabilities.

Looking at the 5 flip row, we have 1 5 10 10 5 1. For 2H, 3H, 4H and 5H we add up the 3rd, 4th, 5th and 6th entries: 10+10+5+1=26.

Summing the whole row, we get 32.

So, the chance of getting at least 2 heads out of 5 flips is 26/32

The probability of getting 1 head should be.
(1/2)^1 x (1/2) ^4 x 5C1
1/32 x (5!/4!1!) = 5/32
Thus,
the probability of getting at least 2 heads should be
P(at least 2 heads) 1 - 5/32 = 27/32

What am I missing here ?

Logitech you are quite correct. Perhaps Stuart Kovinsky from Kaplan can comment on this.

logitech wrote:

Stuart Kovinsky wrote:
Let's look at a much more complicated question:

"A fair coin is flipped 5 times. What's the probability of getting at least 2 heads?"

If we want at least 2 heads, we want 2 heads, 3 heads, 4 heads OR 5 heads. Pretty much whenever we see "OR" in probability, we add the individual probabilities.

Looking at the 5 flip row, we have 1 5 10 10 5 1. For 2H, 3H, 4H and 5H we add up the 3rd, 4th, 5th and 6th entries: 10+10+5+1=26.

Summing the whole row, we get 32.

So, the chance of getting at least 2 heads out of 5 flips is 26/32

Using nCk/2^n

for one HEADS = 5C1/2^5 = 5/32

So for at least two:

1-(5/32) = 27/32

What am I missing here ?

You're missing that we could also get 0 heads, which happens 1/32nd of the time.

So, the probability of getting at least 2H = 1 - Prob(0H or 1H) = 1 - (1/32 + 5/32) = 1 - 6/32 = 26/32 = 13/16.

Not surprisingly, 27/32 is usually one of the answers for this type of question, just to trap the people who forget about the 0H option.

Thank you Stuart!

I love your formula. Thank you for inventing such a convenient formula. However, I don't know why I need to pick the third numbe rof your first formula regarding to the questions of 2 heads in a 3 flips. and so as the second and third questions. Can you please explain to me why you pick the third number but not the second number for 2 heads in 3 flips?

Thank you
mm

mmgmat2008 wrote:I love your formula. Thank you for inventing such a convenient formula. However, I don't know why I need to pick the third numbe rof your first formula regarding to the questions of 2 heads in a 3 flips. and so as the second and third questions. Can you please explain to me why you pick the third number but not the second number for 2 heads in 3 flips?

Thank you
mm

I wish I could take credit for inventing the formula... well heck, I will, just don't tell any math majors! 8)

I assume you're talking about the n=3 row of Pascal's... err Stuart's triangle, which reads:

1 3 3 1

If we want the possibilities for 2 and 3 heads, we have to add the 3rd and 4th numbers in the row because the first number in every row matches 0 of the result you're looking for. You'll see that every row has n+1 entries, i.e. the 3 flip row has 4 entries, the 4 flip row has 5 entries, the 5 flip row has 6 entries, etc...

So, for the 3 row, the 4 entries represent:

1st number: 0 heads
2nd number: 1 head
3rd number: 2 heads
4th number: 3 heads

(just using heads as an example, we could have been counting tails and approached it the exact same way).

Hope that helps!

what is the strategy to fill the numbers in the pascal triangle..
like, if I want to form a row for 7 coin flips, how to I do that?

btw, Great info... thanks a ton

gmataug08 wrote:what is the strategy to fill the numbers in the pascal triangle..
like, if I want to form a row for 7 coin flips, how to I do that?

btw, Great info... thanks a ton

You need to start with a row that you know and work your way down.

If, for example, you remember that the 3 row is 1 3 3 1, you create the next row by starting with 1 on each side and filling in the middle by adding the pairs above. So:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1

is the first 8 rows of the triangle (the first row is n=0). The second number in the row represent the row number (e.g. the second number of the n=7 row is 7).

Icredible tool!!
Thanks