Quantitative Reasoning

I am back with another coin-flip question. :mrgreen:

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

amitdgr wrote:I am back with another coin-flip question. :mrgreen:

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.

WOnderful explanation Stuart:)

"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.

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.

thanks, i was looking at the questions p(0) or p(1) rather than p(2-10). appreciate the help.

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?

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!

Thanks Stuart,

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

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