Assume that boys and girls are equally likely and a couple will have five kids.Find the probability of 1 boy?

Question

Assume that boys and girls are equally likely and a couple will have five kids.Find the probability of 1 boy?

Answer 1

only 1 boy or at least one boy? different answers

exactly one boy  = 5/32

at least one boy = 31/32

its fractions cos its in my head use a calc for percentages but id say roughly

15% and 97%

Answer 2

0girls 5boys

1g 4b

2g 3b

3g 2b

4g 1b

5g 0b

6 options son 1 in 6 or 5to1 or 16.67% for exactly one boy

5 in 6 or 1to5 or 83.33 % for at least one boy.

but to contradict myself the odds of 5 girls are 0.5^5 or 0.5*0.5*0.5*0.5*0.5 which is 0.03125 or 1/32 so agrees with spanther 31/32 for at least one boy.

spanther what is wrong with my original thinking?

only one way to get 5 girls, 5 different orders for one boy 10 for 2.

Answer 3

Let X be the number of male offspring.  X has the binomial distribution with n = 5 trials and success probability p = 0.5



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 = 5 , p = 0.5 )

the mean of the binomial distribution is n * p = 2.5

the variance of the binomial distribution is n * p * (1 - p) = 1.25

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.118034

The Probability Mass Function, PMF,

f(X) = P(X = x) is:

P( X =  0 ) =  0.03125

P( X =  1 ) =  0.15625 <<<< ANSWER

P( X =  2 ) =  0.3125

P( X =  3 ) =  0.3125

P( X =  4 ) =  0.15625

P( X =  5 ) =  0.03125