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Hi I'm sure this is an easy answer but I cant seem to figure this out I am having a tough time understanding the following question:

Using Classical Probability, what is the probability of flipping 3 coins and getting 2 heads and 1 tails.

Thank you,

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    "Concept"? Which concept? – 2017-02-17

2 Answers 2

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So it is the probability to get either : $HHT$, $HTH$ or $THH$.

Note that every throw is independent of the other. Therefore assuming a fair coin you have that $P(HHT) = P(HTH) = P(THH) = 1/8$.

I let you think about why we have this "$1/8$". Note also these 3 events are disjoint.

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    I'm not sure to be honest where the 1/8th comes from – 2017-02-17
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    @EdgarGuzman There is only one way to achieve, for instance, $HHT$, we must get $H$ at first throw which happens with probability $1/2$, then we must get again $H$ which is independent from the first throw, therefore again the probability is $1/2$ and finally we must get $T$ (again independence) so $1/2$ again. Now note that $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$. – 2017-02-17
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    so the answer is 1/8 * 1/8*1/8= .001955 – 2017-02-17
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    @EdgarGuzman You have exactly $1/8$ chance to get precisely $HHT$. However you requirement of having $2$ heads and $1$ tails is less strict. Therefore the probability to fit your requirements will be clearly higher than $1/8 = 0.125$. Namely you'll be happy with either $HHT$, $THT$, $THH$. Therefore the chances **add up** to give you $\frac{3}{8}$ ;) – 2017-02-17
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    Perfect thank you so the answer is 3/8 – 2017-02-17
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The total number of ordered outcomes is $8 = 2 \times 2 \times 2 $ and 3 of these are comprised of 2 H and 1 T so the answer is $\frac{3}{8}$.

Another way of viewing this is as a sequence of independent Bernoulli trials resulting in a Binomial distribution $B(3,\frac{1}{2})$ as you have 3 trials with the probability of success upon each trial = $1/2$. The number of successes (H) here is $X=2$ and number of fails (T) is 1 so the probability is again $P(X=2) = {3 \choose 2}\frac{1}{2}^{2}\frac{1}{2} = \frac{3}{8}$

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    okay, this makes since also, thank you, I was forgetting to include all possible outcomes of each coin. – 2017-02-17