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25 Aug 2006, 17:19
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Hey there. I sit for the GMATic on Tuesday. I am generally pretty strong on probability, but these type of coin toss questions occassionally trip me up:
A coin is tossed four times. How many ways can you get exactly two heads?
Any suggestions on how to best conceptualize this type of question (as well as how you arrived at the answer) would be greatly appreciated.
A coin is tossed four times. How many ways can you get exactly two heads?
Any suggestions on how to best conceptualize this type of question (as well as how you arrived at the answer) would be greatly appreciated.
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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25 Aug 2006, 17:45
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Total outcomes = 2*2*2*2 = 16
Now for 2 heads these are the possibilities
HHTT
THHT
TTHH
HTTH
THTH
HTHT
i.e. 6
Hence answer
= no. of favourable outcomoes/total no. of outcomes
= 6/16
= 3/8
Now for 2 heads these are the possibilities
HHTT
THHT
TTHH
HTTH
THTH
HTHT
i.e. 6
Hence answer
= no. of favourable outcomoes/total no. of outcomes
= 6/16
= 3/8
25 Aug 2006, 20:14
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Such problems can be easily solved by using the binomial formula. For calculating probability of "r" outcomes in an experiment with n-trials, the formula states that:
P = C(n,r) (p^r ) (q^(n-r))
where p is the probability of getting 1 outcome succesful, and q is the probability of getting that same outcome as a failure.
In this case, p = 1/2 , the probability of getting a head in a given coin toss. q = 1/2 as there is no other outcome aside from tail.
Thus: P = C(4.2) (0.5)^2 * (0.5)^2
= 3 (0.25)(0.25) = 3/8
