rtaha2412 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?
5/16
15/32
½
21/32
11/16

Kate will have more than 10$ and less than 15$, if she wins 3 or 4 times.
So, probability of getting 3 tails or 4 tails = 5C3/2^5 + 5C4/2^5 = 10/32 + 5/32 = 15/32.

B

rtaha2412 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?
5/16
15/32
½
21/32
11/16

As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.

When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2
P(5 tails) = 1/32
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.

The correct answer is B.

GMATGuruNY wrote:

rtaha2412 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?
5/16
15/32
½
21/32
11/16

As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.

When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2
P(5 tails) = 1/32
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.

The correct answer is B.

Kate already has $10 . So for the probability that at the end he must have more that $ 10 but less than $ 15 , at least one but not 5 tails should have appeared . So , why aren't we considering 1 and 2 appearances that would in turn earn him 1 or 2 dollars.

Deepthi Subbu wrote:

GMATGuruNY wrote:

rtaha2412 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?
5/16
15/32
½
21/32
11/16

As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.

When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2
P(5 tails) = 1/32
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.

The correct answer is B.

Kate already has $10 . So for the probability that at the end he must have more that $ 10 but less than $ 15 , at least one but not 5 tails should have appeared . So , why aren't we considering 1 and 2 appearances that would in turn earn him 1 or 2 dollars.

For every tails, Kate gets $1; for every heads, she loses $1.

0 tails, 5 heads: Kate has 10+0-5 = $5
1 tails, 4 heads: Kate has 10+1-4 = $7
2 tails, 3 heads: Kate has 10+2-3 = $9
3 tails, 2 heads: Kate has 10+3-2 = $11
4 tails, 1 heads: Kate has 10+4-1 = $13
5 tails, 0 heads: Kate has 10+5-0 = $15

Only 3 tails and 4 tails will give Kate between $10 and $15.

GMATGuruNY wrote:

Deepthi Subbu wrote:

GMATGuruNY wrote:

rtaha2412 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?
5/16
15/32
½
21/32
11/16

As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.

When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2
P(5 tails) = 1/32
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.

The correct answer is B.

Kate already has $10 . So for the probability that at the end he must have more that $ 10 but less than $ 15 , at least one but not 5 tails should have appeared . So , why aren't we considering 1 and 2 appearances that would in turn earn him 1 or 2 dollars.

For every tails, Kate gets $1; for every heads, she loses $1.

0 tails, 5 heads: Kate has 10+0-5 = $5
1 tails, 4 heads: Kate has 10+1-4 = $7
2 tails, 3 heads: Kate has 10+2-3 = $9
3 tails, 2 heads: Kate has 10+3-2 = $11
4 tails, 1 heads: Kate has 10+4-1 = $13
5 tails, 0 heads: Kate has 10+5-0 = $15

Only 3 tails and 4 tails will give Kate between $10 and $15.

Oh ya , got it

GMATGuruNY wrote:

rtaha2412 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?
5/16
15/32
½
21/32
11/16

As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.

When a coin is flipped 5 times, we have an equal chance of getting tails 3, 4 or 5 times as we do of getting tails 0, 1 or 2 times.
Thus, P(3 tails) + P(4 tails) + P(5 tails) = 1/2
P(5 tails) = 1/32
So P(3 tails) + P(4 tails) = 1/2 - 1/32 = 15/32.

The correct answer is B.

Dear GMATGuru
I encountered another solution in which total possibility = 2^5 (which will include a certain number of similar outcomes in different orders)

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:

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 4 wins

My question here:
How we can divide combinations over permutation? as As far as I know, we must either deal with all combinations (in numerator and denominator) or all permutations (in numerator and denominator).

Should not we discard similar outcome from 2^5 to make numerator matches the denominator? or the other way to consider the order in the numerator to match the order in the denominator?

Mo2men wrote:[Dear GMATGuru
I encountered another solution in which total possibility = 2^5 (which will include a certain number of similar outcomes in different orders)

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:

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 4 wins

My question here:
How we can divide combinations over permutation? as As far as I know, we must either deal with all combinations (in numerator and denominator) or all permutations (in numerator and denominator).

3 wins and 2 losses = WWWLL.

One way to determine the number of ways to get 3 wins is to count the number of ways to arrange the 5 letters WWWLL.
The number of ways to arrange 5 elements = 5!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 3! to account for the 3 identical W's and by 2! to account for the 2 identical L's:
5!/(3!2!) = 10

Another approach:
There are 5 positions in the arrangement.
Each position must be occupied by W or L.
A good arrangement occurs when W's occupy A COMBINATION OF 3 POSITIONS in the 5-position arrangement.
Thus, to determine the number of ways to get 3 wins, we can count the number ways to choose a combination of 3 positions from 5 options:
5C3 = 5!/(3!2!) = 10

Each approach yields the same result.