
Reginald J. answered 01/02/24
Personal Probability Professional
Hi there,
This sounds like a Binomial distribution problem considering you have a certain number of "trials' given probabilities and number of successes.
nCx pxqn-x
n=4 (4 kittens, aka trials)
x=3 (take the number of successes to be the number if boys)
p=1/2 (50% chance of "success" (chance of having a boy))
q=1/2 (50% chance of "failure" (chance of having a girl))
4C3 *0.53(0.5)1= 4/16 = 1/4.
I hope this helps!

Reginald J.
I apologize for the grammatical errors in the introduction. There should be a comma before and after "probabilities". * "trials" *number of boys The work is still valid, however.01/02/24