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A fair coin is flipped 6 times. What is the probability of obtaining exactly 2 heads?

Explanation	Calculations	Help
$We note that each coin flip is independent and each outcome is equally likely.$	$P (head) = \frac{1}{2} P (tail) = \frac{1}{2}$	Theory & Tactics Method Card PC1-A Click to view full details
$We determine the total number of possible sequences for six flips.$	$2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$	Theory & Tactics Method Card PC2-A Click to view full details
$We count the number of sequences with exactly two heads by selecting two positions for heads.$	$\frac{6 !}{2 ! ( 6 - 2 )!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 4 \times 3 \times 2 \times 1} = \frac{6 \times 5}{2 \times 1} = 15$	Theory & Tactics Method Card PC3-C Click to view full details
$We find the probability of any specific sequence of six flips as the product of individual probabilities.$	$(\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{64}$	Theory & Tactics Method Card PC1-C Click to view full details
$We multiply the number of favorable sequences by the probability of any specific sequence.$	$15 \times \frac{1}{64} = \frac{15}{64}$

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A

A fair coin is flipped 6 times. What is the probability of obtaining exactly 2 heads?

Explanation	Calculations	Help
$We note that each coin flip is independent and each outcome is equally likely.$	$P (head) = \frac{1}{2} P (tail) = \frac{1}{2}$	Theory & Tactics Method Card PC1-A Click to view full details
$We determine the total number of possible sequences for six flips.$	$2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$	Theory & Tactics Method Card PC2-A Click to view full details
$We count the number of sequences with exactly two heads by selecting two positions for heads.$	$\frac{6 !}{2 ! ( 6 - 2 )!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 4 \times 3 \times 2 \times 1} = \frac{6 \times 5}{2 \times 1} = 15$	Theory & Tactics Method Card PC3-C Click to view full details
$We find the probability of any specific sequence of six flips as the product of individual probabilities.$	$(\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{64}$	Theory & Tactics Method Card PC1-C Click to view full details
$We multiply the number of favorable sequences by the probability of any specific sequence.$	$15 \times \frac{1}{64} = \frac{15}{64}$

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A