Questions about the world of GMAT Math from other sources and general math related questions.
goelmohit2002
Students
 
Posts: 226
Joined: Sat Jul 04, 2009 8:40 am
 

Head-Tail Question

by goelmohit2002 Fri Aug 07, 2009 1:55 am

Hi All,

Probability of getting a head on a toss of non fair coin is 0.6. In 6 tosses of a coin, what is the probability of getting atleast 5 heads ?

OA = Will post after some discussion.

Thanks
Mohit
sharok50
Students
 
Posts: 22
Joined: Fri Mar 20, 2009 10:24 pm
 

Re: Head-Tail Question

by sharok50 Wed Aug 12, 2009 12:30 pm

nCr X p^r X q^n-r

=> 6C5 X (.6)^5 X (.4)^6-5

i hope this is the way.
goelmohit2002
Students
 
Posts: 226
Joined: Sat Jul 04, 2009 8:40 am
 

Re: Head-Tail Question

by goelmohit2002 Wed Aug 12, 2009 1:20 pm

sharok50 Wrote:nCr X p^r X q^n-r

=> 6C5 X (.6)^5 X (.4)^6-5

i hope this is the way.


Yes thanks. The above is the OA.
Ben Ku
ManhattanGMAT Staff
 
Posts: 817
Joined: Sat Nov 03, 2007 7:49 pm
 

Re: Head-Tail Question

by Ben Ku Tue Aug 18, 2009 9:27 pm

sharok50's response only takes into account the probability of tossing exactly 5 heads in 6 tosses. However, it does not add in the probability of tossing 6 heads.

I look at the problem this way:
P(5 Heads) + P(6 Heads)

P(6 Heads) is easy. Each toss has a 0.6 chance of being heads, so P(6 Heads) = (0.6)^6.

P(5 Heads) is a bit more complicated, because there are different ways to get 5 heads. One combination of getting 5 Heads is HHHHHT. The probability of this combination is (0.6)(0.6)(0.6)(0.6)(0.6)(0.4) = (0.6)^5(0.4). [The 0.4 is there because that's the probability of getting tails.] Using the anagram method, there are 6!/5! = 6 different combinations of getting 5 Heads. So P(5 Heads) = (6)(0.6^5)(0.4) = (4)(0.6^6)

P(5 Heads) + P(6 Heads) = (4)(0.6^6) + (0.6^6) = 5(0.6^6)
Ben Ku
Instructor
ManhattanGMAT