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# This topic has 24 expert replies and 45 member replies

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KristenH88 Junior | Next Rank: 30 Posts
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Wed Nov 12, 2014 6:16 pm
This is AMAZING. You should teach the whole test on here. It made it sooooo much simpler on each problem I did. Thank you for explaining, I definitely didn't understand how 4C3 turned in to 4. Now I get it. THANK YOU!

mariolita Newbie | Next Rank: 10 Posts
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Tue Dec 16, 2014 2:27 am
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

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Rich.C@EMPOWERgmat.com Elite Legendary Member
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Tue Dec 16, 2014 10:29 am
Hi mariolita,

If you can post the question here (either take a picture or transcribe it), then I'm sure you'll get some responses.

GMAT assassins aren't born, they're made,
Rich

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Brent@GMATPrepNow GMAT Instructor
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Tue Dec 16, 2014 3:24 pm
mariolita wrote:
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
I believe this is the question you're referring to:

Quote:
The probability is ½ 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
Here's one approach:

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 tails) = 1 - P(not getting at least 1 tails)
What does it mean to not get at least 1 tails? It means getting zero tails.
So, we can write: P(getting at least 1 tails) = 1 - P(getting zero tails)

Now let's calculate P(getting zero tails)
What needs to happen in order to get zero tails?
Well, we need heads on the first toss and heads on the second toss and heads on the third toss.
We can write P(getting zero tails) = P(heads on 1st AND heads on 2nd AND heads on 3rd)
This means that P(getting zero tails) = P(heads on 1st) x P(heads on 2nd) x P(heads on 3rd)
Which means P(getting zero tails) = (1/2)x(1/2)x(1/2)= 1/8

We're now ready to answer the question.
P(getting at least 1 tails) = 1 - P(not getting at least 1 tails)
= 1 - 1/8
= 7/8
= D

Cheers,
Brent

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Brent@GMATPrepNow GMAT Instructor
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Tue Dec 16, 2014 3:24 pm
Quote:
The probability is ½ 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
Another approach is to draw a TREE DIAGRAM that shows all of the possible outcomes.

Here's the 1st toss:

Then add the 2nd toss:

Then the 3rd:

So, as you can see there are 8 possible outcomes:

Now let's place a checkmark beside those outcomes that satisfy the condition of having AT LEAST one tails.

As we can see, 7 of the 8 outcomes satisfy this condition. So, the probability is 7/8

Answer: D

Cheers,
Brent

_________________
Brent Hanneson – Creator of GMATPrepNow.com
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ocina Newbie | Next Rank: 10 Posts
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Fri Feb 20, 2015 3:01 am
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. ocina Newbie | Next Rank: 10 Posts Joined 20 Feb 2015 Posted: 3 messages Fri Feb 20, 2015 3:01 am 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.
блиндирани врати врати входни врати

Last edited by ocina on Wed Feb 25, 2015 6:58 am; edited 1 time in total

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Rich.C@EMPOWERgmat.com Elite Legendary Member
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Fri Feb 20, 2015 4:25 pm
Hi ocina,

This is a layered probability question that requires some additional math (the combination formula, or the equivalent "mapping" of all possibilities).

We're told that the coin will be flipped 5 times; with every "win", Kate gets $1 and with every "loss", Kate loses$1. Kate starts with $10. We're asked for the probability that Kate ends up with MORE than$10 but less than $15 after 5 tosses. Let's start with the total number of possible outcomes. Since each coin has 2 options (heads or tails), there are 2^5 = 32 possible outcomes for the 5 flips (which will include a certain number of similar outcomes in different orders - for example HHTTT and THTHT). To end up with MORE than$10, Kate has to win MORE tosses than she loses. However, if she were to win all 5 tosses, she'd have $15 and we want her to end up with LESS than$15. This means that Kate has to win EITHER 3 times or 4 times.

Since it does not matter which of the 5 tosses is won, as long as it's either 3 or 4 of them, we can use the combination formula:

Combinations = N!/[K!(N-K)!]

For 3 wins, we have 5!/3!2! = 10 possible combinations of 3 wins

For 4 wins, we have 5!/4!1! = 5 possible combinations of 3 wins

So, there are a total of 10+5 = 15 combinations of 5 tosses that "fit" what we're looking for and 32 possible outcomes total.

Final Answer: 15/32

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

billhenry13 Newbie | Next Rank: 10 Posts
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Tue Oct 24, 2017 3:44 am
The possibility of throwing a coin here is explored with simulation. When asked about the question, what is the probability that a coin toss will come face to face, most people respond without hesitation that it is 50%, 1/2 or 0.5 https://www.vcetests.com/2V0-620-vce.html
We get this probability on the assumption that the currency is fair, or that the heads and tail are equal

The potential for both potential outcomes is:

The number of results in the event العدد The total number of possible outcomes

For currency, the number of results for heads = 1

Total number of possible results = 2

So, we get 1/2

However, if you suspect that the currency may not be fair, you can throw a coin a large number of times and promised a number of heads

Suppose you throw a 100 coin and you get 60 heads, then you know that the best estimate for a head is 60/100 = 0.6 https://www.vcetests.com/2V0-621-vce.html

This view is called the possibility of estimating the relative frequency of probability

The interesting thing about this is that the more you throw a coin, the closer it is to 0.5[url][/url][url]

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Matt@VeritasPrep GMAT Instructor
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Thu Nov 09, 2017 6:06 pm
Zombie bots run amok! What's the probability that another one rises from the digital earth to haunt this thread?

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