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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.

Coin is flipped 5 times hence the sample space contains 2^5 ie 32 outcomes. We get this by general formula number of outcomes ^ number of repetitions. Now Kate can have more than 10 and less than 15 only if there are 4 Tails and 1 Head or 3 Tails and 2 Heads, this can be found out by simple manual calculation as below: All 5 Tails, Kate: 15 David: 5 4 Tails, 1 Head, Kate: 13 David: 7 3 Tails, 2 Head, Kate: 11 David: 9 less than 2 tails then Kates total goes below 10 which is not to be considered.

Now cases with 4 tails and 1 head are 5!/4!*1! = 5 --> Simple arrangement rule Now cases with 3 tails and 2 heads are 5!/3!*2! = 10 --> Simple arrangement rule

I answered this problem logically rather than using a formula.

With 5 coin tosses, there is a 50% chance Kate will have more than $10, however the problem states what is the possibility of having more than $10 but less than $15. In order for Kate to have $15, she would need to win all 5 tosses (1 out of 32 possibility).

So...1/2 - 1/32 = 15/32.

Even if you couldn't figure out the 1/32 possibility, 15/32 is the only answer just below 1/2. None of the other answers are even close.

If this is confusing just ignore it. I think in words more than symbols.

After reading the given answer I see in this question that it is the correct answer. However, I can't seem to remember the notation for these kind of questions. So I am trying to assemble a method that I can remember.

I am thinking something like this

\frac{Ways to Satisfied Conditions}{Possible Outcomes}=Probability of Conditions

For a question where multiple outcomes satisfy the conditions. Something like this might be better.

\frac{Ways to GetOutcome#1 + Ways to Get Outcome #2}{Possible Outcomes}

For this question I write \frac{Different Ways To Get 4 Wins For Kate+Different Ways To Get 3 Wins For Kate}{Possible Outcomes} then \frac{10+5}{32}=15/32

And for those who want a further challenge, I suggest this question

A dice is rolled 5 times, what are the chances of rolling a 6, at least 3 times but not more than 4 times?

Only in 2 cases Kate will have more than $10 and less than $15. Either 3 heads/2 tails or 4 heads/1 tail Probability of having 3h and 2t is 5c3*(1/32)=10/32 Probability of having 4h and 1t is 5c4*(1/32)=5/32 So probability of either of the above is 10/32 + 5/32 = 15/32 (Ans) _________________

Re: m05 Q17 Probability/Combination [#permalink]
01 Aug 2012, 23:49

1

This post received KUDOS

Expert's post

dpvtank wrote:

GZR4DR wrote:

I answered this problem logically rather than using a formula.

With 5 coin tosses, there is a 50% chance Kate will have more than $10, however the problem states what is the possibility of having more than $10 but less than $15. In order for Kate to have $15, she would need to win all 5 tosses (1 out of 32 possibility).

So...1/2 - 1/32 = 15/32.

Even if you couldn't figure out the 1/32 possibility, 15/32 is the only answer just below 1/2. None of the other answers are even close.

This is kind of a silly question: can you or someone please explain how there is a 50% chance that Kate will have more than $10?

I get the \frac{1}{32} part for her winning all 5 tosses.

After each toss Kate either gains $1 or losses $1.

After 5 tosses she'll have either more than $10 or less than $10 (she cannot have exactly $10). Since there is no reason one case to have more chances to appear than another then the probability that Kate will have more than $10 is 1/2 and the probability that Kate will have less than $10 is 1/2 too.

Hi. Think about it this way. If K wins only 1 time that means she loses 4 time (the coin is flipped 5 times in total). So this way K will have

10 + 1 - 4 = 7 dollars (less than we need)

If K wins 2 time (which means she loses 3 times), she will have

10 + 2 - 3 = 9 dollars (less than we need)

If K wins all 5 times she'll end up with $15 (more than we need).

Now you see why we check only the combinations when K wins 3 (K will have $11) and 4 (K will have $13) times. Hope this helps.

d2touge wrote:

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

So to answer this, we find the total combinations when you flip the coin 5 times. --> 2^5 = 32.

The explanation tells us to find the combinations when kate wins 3 times and 4 times. --> 5C3 and 5C4. we get 10 and 5 respectively.

Lastly, we simply find the probability (5/32) + (10/32) = 15/32. and that is the answer..

***Why do we only find the combinations when Kate wins 3 times and 4 times? Why not find 1 and 2 times?

Re: m05 Q17 Probability/Combination [#permalink]
04 Jun 2010, 01:21

You may want to look at it this way.

How many outcomes are there when you flip a coin once? There are 2 outcomes (Heads or Tails). If you flip a coin twice? - 4 outcomes (2^2). If you flip a coin twice, you can get these 4 outcomes: HH, HT, TH, TT. There are no other outcomes than these 4. When you flip that coin three times, you will get 8 outcomes (2^3). Here they are: HHH, HHT, HTT, TTT, TTH, THH, HTH, THT. So, flipping a coin 5 times gives 32 = 2^5 outcomes.

You might want to take a look at this Probability thread from our GMAT Math Book: math-probability-87244.html

There are two winning scenarios. Kate should win either 3 or 4 times.

This is how I solved it: scenarios 1, Kate wins 4 times = 3*(1/2)^3 + (1/2)^2 = 5/8 scenarios 2, Kate wins 3 times = 4*(1/2)^4+ (1/2) = 3/4 5/8*3/4 = 15/32

I believe I solved it in a stupid way but that somehow worked. Can someone tell whether I just guessed or this method works?

There are six combinations, so we use the sixth row. Or another angle to look at it is there are five flips, and then we count the first row as the zero row. Either way we arrive at the 1-5-10-10-5-1.

Now write out the combinations and money Kate gets.

Re: m05 Q17 Probability/Combination [#permalink]
01 Aug 2012, 19:48

GZR4DR wrote:

I answered this problem logically rather than using a formula.

With 5 coin tosses, there is a 50% chance Kate will have more than $10, however the problem states what is the possibility of having more than $10 but less than $15. In order for Kate to have $15, she would need to win all 5 tosses (1 out of 32 possibility).

So...1/2 - 1/32 = 15/32.

Even if you couldn't figure out the 1/32 possibility, 15/32 is the only answer just below 1/2. None of the other answers are even close.

This is kind of a silly question: can you or someone please explain how there is a 50% chance that Kate will have more than $10?

I get the \frac{1}{32} part for her winning all 5 tosses.

Re: m05 Q17 Probability/Combination [#permalink]
18 Nov 2012, 00:42

1

This post was BOOKMARKED

Approach 1: Find the total number of combinations = 2^5=32. Find the number of combinations when Kate wins: out of 5 games, she can win 3 or 4 times only as 5 victories would put her over the $14 mark and less than 3 victories, below the $10.