Too many puppies

Probability Level 2

You have a dog who is equally likely to give birth to a male dog or a female dog.

Supposing it has four off springs, which of the following is the most likely split?

Trust me, I do not like dogs except that puppy linux was my all time favorite.

Four female or Four male Two female, Two male (Three female, One male) or (One female, Three Male)

Agnishom Chattopadhyay by Agnishom Chattopadhyay , 22, India

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4 solutions

Bobbym None
Mar 2, 2015

Easily done with a probability generating function.

( m 2 + f 2 ) 4 = f 4 16 + f 3 \ m 4 + 3 f 2 m 2 8 + f m 3 4 + m 4 16 \left(\frac{m}{2}+\frac{f}{2}\right)^4=\frac{f^4}{16}+\frac{f^3 \ m}{4}+\frac{3 f^2 m^2}{8}+\frac{f m^3}{4}+\frac{m^4}{16}

m 3 f + f 3 m = 1 2 m^3f + f^3m = \frac{1}{2} which is the probability of 3 males or 3 females

m 2 f 2 = 3 8 m^2f^2 = \frac{3}{8} which is the probability of 2 of each.

m 4 + f 4 = 1 8 m^4 + f^4 = \frac{1}{8} which is the probability of 4 of each.

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Venture Hi
Mar 2, 2015

The sample space is ( m,m,m,m),(f,f,f,f),(m,f,f,f),(m,m,f,f),(m,m,m,f). The chance of getting option A and B is 1/5 whilst the chance for optiocn C is 2/5.

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The sample space has to take into account the permutations of the outcomes you have listed. The full sample space has 2 4 = 16 2^{4} = 16 possible outcomes, with P ( ( 1 , 3 ) , ( 3 , 1 ) ) = 8 16 , P ( ( 0 , 4 ) , ( 4 , 0 ) ) = 2 16 P((1,3),(3,1)) = \dfrac{8}{16}, P((0,4),(4,0)) = \dfrac{2}{16} and P ( ( 2 , 2 ) ) = 6 16 . P((2,2)) = \dfrac{6}{16}.

Brian Charlesworth - 6 years, 3 months ago

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yes but the question ask us to pick one out of the three options..with the one having the highest chance of happening.

Venture HI - 6 years, 3 months ago

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Yes, I agree, and given the three options, the one with either 3 males, 1 female or 1 male, 3 females has a probability of 1 2 \dfrac{1}{2} of occurring, which is the highest of the three.

Brian Charlesworth - 6 years, 3 months ago

I don't think that is correct

Agnishom Chattopadhyay - 6 years, 3 months ago
Rajath Naik
Mar 2, 2015

The answer is quite simple.we no that the total number of pups=4.so the totalnumber of permutations out of male and female pups to be arranged in 4 differnet positions is (²C1)(²C1)(²C1)(²C1)=16.so there are 16 different ways in which the linux puppy can venture to reproduce.the first option is the least likeliest as all males and all females accounts to only 2/16 of the probability.and for the third option I tried trial and error as it was easier..it accounts for 4/16 of the probability.which is still not great.so the option 2 which accounts for the rest of the probability I.e 10/16 should be most likely...

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The chance of having all male = ( 1 2 ) 4 (\frac{1}{2})^4 = chance of all female.

Therefore, chance of all male or all female = 1 8 \frac{1}{8}

Chance of two male and two female = C ( 4 , 2 ) ∗ ( 1 2 ) 4 = 3 8 C(4,2)*(\frac{1}{2})^4 = \frac{3}{8}

Chance of one female and three males or vice versa = 1 − 1 8 − 3 8 = 1 2 1-\frac{1}{8}-\frac{3}{8} = \frac{1}{2}

The third option has the greatest chance.

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