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# probability question - coin toss

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Intern
Joined: 16 Jul 2006
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probability question - coin toss [#permalink]

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25 Aug 2006, 17:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.
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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
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Re: probability question - coin toss [#permalink]

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25 Aug 2006, 20:14
pws wrote:
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.

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
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Re: probability question - coin toss [#permalink]

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25 Aug 2006, 20:24
pws wrote:
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.

Use binomial distribution, when the possible favourable/nonfavourable events are fixed 3/8
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26 Aug 2006, 07:23
Agree that the binomial distribution is the way to go here:

P = nCr X p^r X q^(n-r)

will give the probability. However, this does not give us the actual number of ways for getting 2H and 2T. To get that we multiply the probability by the total number of outcomes.

Hence we get 6/16 X 2^4 = 6
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26 Aug 2006, 15:30
HI guys....

I have a question

How many ways can you get exactly two heads? => this question means to calculate probability?

or means 4P2?

I will really appreciate your answers---
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Woody crack the GMAT

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26 Aug 2006, 21:50
zorro 13 has raised an interesting question..

How do we know if it is the probability or if it literally means the number of ways 4C2 that you could get exactly two heads?

I think the answer choices should give a clue.
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26 Aug 2006, 22:47
Read my answer again. The question asks for the number of ways. Once you know the probability (which is easier to calculate in this case), you can multiply it by the total number of possibilities to get the needed answer. Does this help?
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26 Aug 2006, 22:52
yep, in this case, it is the number of ways not the probability
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27 Aug 2006, 00:27
Thank you guys.....

.
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Woody crack the GMAT

27 Aug 2006, 00:27
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# probability question - coin toss

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