Which are removable discontinuities?
1 answer:
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Answer: There are 4 male goats.
Step-by-step explanation:
We know that n of the 10 goats are male.
The probability that in a random selection, the selected goat is a male, is equal to the quotient between the number of male goats (n) and the total number of goats (10)
The probability is;
p = n/10
Now the total number of goats is 9, and the number of male goats is n -1
then the probability of selecting a male goat again is:
q = (n-1)/9
The joint probability (the probability that the two selected goats are male) is equal to the product of the individual probabilities, this is
P = p*q = (n/10)*((n-1)/9)
And we know that this probability is equal to 2/15
Then we have:
(n/10)*((n-1)/9) = 2/15
(n*(n-1))/90 = 2/15
n*(n-1) = 90*2/15 = 12
n^2 - n = 12
n^2 - n - 12 = 0
This is a quadratic equation, we can find the solutions if we use Bhaskara's formula:
For an equation:
a*x^2 + b*x + c = 0
The two solutions are given by:
For our case, the solutions will be:
The two solutions are:
n = (1 - 7)/2 = -3 (this solution does not make sense, we can not have a negative number of goats)
The other solution is:
n = (1 + 7)/2 = 4
This solution does make sense, this means that we have 4 male goats.