Genetics and Probability Questions?

Question

1) Indicate the probability of each of the following:
a) 4 girls
b) 3 girls, 1 boy
c) 2 girls, 2 boys
d) 1 girl, 3 boys
e) 4 boys

2) What would be the porbability of 3 girls and 3 boys being born on the same day?

3) If AaBb is mated toAaB, what is the porbability that the offspring will be each of the following:

a) either AABb or AaBB?
b) either AaBb or aaBb?
c) either Aabb or aaBB?
d) either AABB or aaBB?
e) either phenotype A-B-or phenotype aaB-?

Merlyn · Accepted Answer

1)

Let X be the number of girls born.  X has the binomial distribution with n = 4 trials and success probability p = 0.5, assuming the probability of male/female is equal.
 
In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.

The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.
Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.

X ~ Binomial( n = 4 , p = 0.5 )

the mean of the binomial distribution is n * p = 2 
the variance of the binomial distribution is n * p * (1 - p) = 1 
the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1 

The Probability Mass Function, PMF,
 f(X) = P(X = x) is:

P( X =  0 ) =  0.0625 <- zero girls, 4 boys
P( X =  1 ) =  0.25 <- 1 girl, 3 boys
P( X =  2 ) =  0.375 <- 2 girls, 2 boys
P( X =  3 ) =  0.25 <- 3 girls, 1 boys
P( X =  4 ) =  0.0625 <- 4 girls, 0 boys



2)
X ~ Binomial( n = 6 , p = 0.5 )
P( X =  3 ) =  0.3125

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Genetics and Probability Questions?

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