A
B
C
D
E
Kate will have more than 10$ and less than 15$, if she wins 3 or 4 times.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.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 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.GMATGuruNY wrote:As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.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
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.
For every tails, Kate gets $1; for every heads, she loses $1.Deepthi Subbu wrote: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.GMATGuruNY wrote:As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.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
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.
Oh ya , got itGMATGuruNY wrote:For every tails, Kate gets $1; for every heads, she loses $1.Deepthi Subbu wrote: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.GMATGuruNY wrote:As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.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
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.
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.
Dear GMATGuruGMATGuruNY wrote:As noted above, Kate will have between $10 and $15 if the coin lands on tails 3 or 4 times.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
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.
3 wins and 2 losses = WWWLL.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).
