Answer:
[tex]\frac{^3C_1\times ^4C_1}{^{12}C_2}[/tex]
Step-by-step explanation:
Given,
Art electives = 3,
History electives = 4,
Computer electives = 5,
Total number of electives = 3 + 4 + 5 = 12,
Since, if a student chooses an art elective and a history elective,
So, the total combination of choosing an art elective and a history elective = [tex]^3C_1\times ^4C_1[/tex]
Also, the total combination of choosing any 2 subjects out of 12 subjects = [tex]^{12}C_2[/tex]
Hence, the probability that a student chooses an art elective and a history elective = [tex]\frac{\text{Total combination of choosing an art elective and a history}}{\text{ Total combination of choosing any 2 subjects}}[/tex]
[tex]=\frac{^3C_1\times ^4C_1}{^{12}C_2}[/tex]
Which is the required expression.
Answer: Hello!
we have:
3 art electives
4 history electives
5 computer electives
which adds to a total of 12.
If the selection is random, each elective has the same probability.
The probability of selecting an art electives is the quotient between the number of art electives and the total number of electives:
3/12
suppose that this event is true, now we need to see the probability of choosing also a history elective;
We do the same process as before, we have 4 history electives and, because we already selected 1 in the previous step, we have a total of 11 electives:
the probability now is 4/11.
Now we want to calculate the joint probability of bot events is equal to the product of their probabilities; this is:
p= (3/12)*(4/12) = (4*3)/(11*12) = 12/(11*12) = 1/11
But there is also the case where the selection is in the other order (first history and second art) so the probability is equal to
2*1/11 = 2/11
