A litter of eight puppies consists of 5 males and 3 females. Two puppies are chosen at random, without replacement. What is the probability that both puppies will be female?
prob(your stated event) = C(3,2)/C(8,2) = 3/28
btw, breakdown of probabilities:
both females = 3/28 (see above)
both males = C(5,2)/C(8,2) = 10/28
one male, one female = C(5,1)*C(3,1)/C(8,2 = 15/28
note 3/28 + 10/28 + 15/28 = 28/28 = 1
To find the probability that both puppies will be female, you need to consider the total number of possible outcomes and the number of favorable outcomes.
First, let's find the total number of possible outcomes. We have a litter of 8 puppies, and we are choosing 2 of them. So, we can use the combination formula:
nCr = n! / (r!(n - r)!)
Here, n is the total number of puppies (8) and r is the number of puppies chosen (2). Plugging in the values, we get:
8C2 = 8! / (2!(8 - 2)!) = (8 * 7) / (2 * 1) = 28
So, there are 28 possible outcomes when choosing 2 puppies from a litter of 8.
Now, let's find the number of favorable outcomes. We want to choose 2 female puppies from the litter, and there are 3 females in total. We can use the combination formula again:
3C2 = 3! / (2!(3 - 2)!) = (3 * 2) / (2 * 1) = 3
So, there are 3 favorable outcomes when choosing 2 female puppies from a litter of 3 females.
Finally, we can find the probability of both puppies being female by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Favorable outcomes / Total outcomes
= 3 / 28
Therefore, the probability that both puppies will be female is 3/28 or approximately 0.107.