A family plans to adopt 5 puppies. Each puppy is e... | Free GMAT Practice Question | GMAT Panda

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A family plans to adopt 5 puppies. Each puppy is equally likely to be male or female. If they adopt exactly 5 puppies, what is the probability that exactly 3 of them are male and 2 are female?

Explanation	Calculations	Help
$We note that each puppy's gender is independent of the others and each puppy has equal chance to be male or female.$	$P (male) = \frac{1}{2} P (female) = \frac{1}{2}$	Theory & Tactics Method Card PC1-A Click to view full details
$We determine the total number of possible gender sequences in five adoptions by multiplying the number of outcomes for each adoption.$	$2 \times 2 \times 2 \times 2 \times 2 = 32$	Theory & Tactics Method Card PC2-A Click to view full details
$We count how many sequences have exactly three males by selecting three positions out of five.$	$\frac{5 !}{3 ! ( 5 - 3 )!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 2 \times 1} = \frac{5 \times 4}{2 \times 1} = 10$	Theory & Tactics Method Card PC3-C Click to view full details
$We find the probability of any specific sequence of five adoptions as the product of individual probabilities since adoptions are independent.$	$(\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{32}$	Theory & Tactics Method Card PC1-C Click to view full details
$We divide the number of favorable sequences by the total sequences and then simplify the fraction.$	$\frac{10}{32} = \frac{2 \times 5}{2 \times 16} = \frac{5}{16}$	Theory & Tactics Method Card FDPR2-A Click to view full details

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B

A family plans to adopt 5 puppies. Each puppy is equally likely to be male or female. If they adopt exactly 5 puppies, what is the probability that exactly 3 of them are male and 2 are female?

Explanation	Calculations	Help
$We note that each puppy's gender is independent of the others and each puppy has equal chance to be male or female.$	$P (male) = \frac{1}{2} P (female) = \frac{1}{2}$	Theory & Tactics Method Card PC1-A Click to view full details
$We determine the total number of possible gender sequences in five adoptions by multiplying the number of outcomes for each adoption.$	$2 \times 2 \times 2 \times 2 \times 2 = 32$	Theory & Tactics Method Card PC2-A Click to view full details
$We count how many sequences have exactly three males by selecting three positions out of five.$	$\frac{5 !}{3 ! ( 5 - 3 )!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 2 \times 1} = \frac{5 \times 4}{2 \times 1} = 10$	Theory & Tactics Method Card PC3-C Click to view full details
$We find the probability of any specific sequence of five adoptions as the product of individual probabilities since adoptions are independent.$	$(\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{32}$	Theory & Tactics Method Card PC1-C Click to view full details
$We divide the number of favorable sequences by the total sequences and then simplify the fraction.$	$\frac{10}{32} = \frac{2 \times 5}{2 \times 16} = \frac{5}{16}$	Theory & Tactics Method Card FDPR2-A Click to view full details

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B