Suppose a fair coin is tossed $6$ times. What is t... | Free GMAT Practice Question | GMAT Panda

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Suppose a fair coin is tossed 6 times. What is the probability that it lands heads exactly 3 times and tails exactly 3 times?

Explanation	Calculations	Help
$We note that each coin toss is independent and each toss has an equal chance of landing heads or tails.$	$P (heads) = \frac{1}{2} P (tails) = \frac{1}{2}$	Theory & Tactics Method Card PC1-A Click to view full details
$We determine the total number of possible sequences of outcomes in six tosses by multiplying the number of outcomes for each toss.$	$2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$	Theory & Tactics Method Card PC2-A Click to view full details
$We count how many sequences have exactly three heads by selecting three positions out of six.$	$\frac{6 !}{3 ! 3 !} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 3 \times 2 \times 1} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$	Theory & Tactics Method Card PC3-C Click to view full details
$We find the probability of any specific sequence of three heads and three tails as the product of individual probabilities since the tosses are independent.$	$(\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{64}$	Theory & Tactics Method Card PC1-C Click to view full details
$We divide the number of favorable sequences by the total sequences and then simplify the fraction.$	$\frac{20}{64} = \frac{4 \times 5}{4 \times 16} = \frac{5}{16}$	Theory & Tactics Method Card FDPR2-A Click to view full details

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A

Suppose a fair coin is tossed 6 times. What is the probability that it lands heads exactly 3 times and tails exactly 3 times?

Explanation	Calculations	Help
$We note that each coin toss is independent and each toss has an equal chance of landing heads or tails.$	$P (heads) = \frac{1}{2} P (tails) = \frac{1}{2}$	Theory & Tactics Method Card PC1-A Click to view full details
$We determine the total number of possible sequences of outcomes in six tosses by multiplying the number of outcomes for each toss.$	$2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$	Theory & Tactics Method Card PC2-A Click to view full details
$We count how many sequences have exactly three heads by selecting three positions out of six.$	$\frac{6 !}{3 ! 3 !} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 3 \times 2 \times 1} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$	Theory & Tactics Method Card PC3-C Click to view full details
$We find the probability of any specific sequence of three heads and three tails as the product of individual probabilities since the tosses are independent.$	$(\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{64}$	Theory & Tactics Method Card PC1-C Click to view full details
$We divide the number of favorable sequences by the total sequences and then simplify the fraction.$	$\frac{20}{64} = \frac{4 \times 5}{4 \times 16} = \frac{5}{16}$	Theory & Tactics Method Card FDPR2-A Click to view full details

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A