Probability - Kate & Danny

Expert replies

cypherskull: Senior | Next Rank: 100 Posts; Posts: 94; Joined: Sat Mar 31, 2012 3:39 am; Location: Calcutta; Thanked: 8 times

by cypherskull » Sat Aug 25, 2012 7:56 am

Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability that Kate has more than $10 but less than $15?

a.5/16
b.1/2
c.12/30
d.15/32
e.3/8

I'm invariably making the same mistake in probability questions over and over. Yet, I'm never really sure where I'm going wrong. This is how I approached the problem and got an answer which is nowhere in the options

Kate wins all 5 = Kate ends with $15
Kate wins 4 & Danny wins 1 = Kate ends with $13
Kate wins 3 & Danny wins 2 = Kate ends with $11
Kate wins 2 & Danny wins 3 = Kate ends with $9
Kate wins 1 & Danny wins 4 = Kate ends with $7
Kate wins 0 & Danny wins 5 = Kate ends with $5

Reqd probability = Desired results (in bold face)/All possibilities = 2/6 = 1/3

Obviously, I'm wrong! Experts please help me with the required concepts here!!

Regards,
Sunit

________________________________

Kill all my demons..And my angels might die too!

Source: Beat The GMAT — Problem Solving | Sat Aug 25, 2012 7:56 am

GaneshMalkar: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Tue Jul 31, 2012 6:53 amFollowed by: 1 member; Thanked: 8 times

by GaneshMalkar » Sat Aug 25, 2012 10:39 am

cypherskull wrote:Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability that Kate has more than $10 but less than $15?

a.5/16
b.1/2
c.12/30
d.15/32
e.3/8

I'm invariably making the same mistake in probability questions over and over. Yet, I'm never really sure where I'm going wrong. This is how I approached the problem and got an answer which is nowhere in the options

Kate wins all 5 = Kate ends with $15
Kate wins 4 & Danny wins 1 = Kate ends with $13
Kate wins 3 & Danny wins 2 = Kate ends with $11
Kate wins 2 & Danny wins 3 = Kate ends with $9
Kate wins 1 & Danny wins 4 = Kate ends with $7
Kate wins 0 & Danny wins 5 = Kate ends with $5

Reqd probability = Desired results (in bold face)/All possibilities = 2/6 = 1/3

Obviously, I'm wrong! Experts please help me with the required concepts here!!

The order of the head or tail matters as it is stick to money

Its a permutation problem!!!

Total number of possibilities = 2^5

Now to get kate more than 10 and less than 15, kate has to win 4 or 3 times....

Probability of getting 4 wins = P(4) = 5C4 / 2^5 = 5/32
Probability of getting 3 wins = P(3) = 5C3 / 2^5 = 10/32

Total probability = 15/32

If you cant explain it simply you dont understand it well enough!!!
- Genius

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Aug 25, 2012 6:18 pm

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