Answer:
The chances Gavis get four or more correct problems is 8/11 or 72.72%
Step-by-step explanation:
The exam is composed of 6 problems out of 12 possible cases (Pc=12). There are 2 groups of problems:
The 8 problems that Gavin has the answer (Problems A).
The 4 problems that Gavin hasn´t the answer (problems B).
Therefore:
P(A≥4)= P(A=4) ∪ P(A=5) ∪ P(A=6) = P(A=4) + P(A=5) + P(A=6)
Before we start analyzing the problem, we have to understand that problems in the exam are selected at random, but a problem can´t be selected twice. therefore picking a specific problem will reduce the pool of that specific group and of the total number of available problems.
If we call [tex]X_{ij}[/tex] to the probability of an answer of the X group to be the i° picked problem from the j° picked problem of that group:
[tex]X_{ij}=\frac{N_{GRX}-j+1}{12-i+1}[/tex] with [tex]N_{GRX}[/tex] the total number of problems in that group.
We analyze now 3 different problems:
[tex]P(A=6)=A_{11} \cdot A_{22} \cdot A_{33} \cdot A_{44} \cdot A_{55} \cdot A_{66}\\P(A=6)=\frac{8}{12} \frac{7}{11} \frac{6}{10} \frac{5}{9} \frac{4}{8} \frac{3}{7} = \frac{1}{33}[/tex]
[tex]P(A=5)=A_{11} \cdot A_{22} \cdot A_{33} \cdot A_{44} \cdot A_{55} \cdot B_{61}+ A_{11} \cdot A_{22} \cdot A_{33} \cdot A_{44} \cdot B_{51} \cdot A_{65}+...+B_{11} \cdot A_{21} \cdot A_{32} \cdot A_{43} \cdot A_{54} \cdot A_{65}\\[/tex]
[tex]P(A=5)=\frac{8}{12} \frac{7}{11} \frac{6}{10} \frac{5}{9} \frac{4}{8} \frac{4}{7} + \frac{8}{12} \frac{7}{11} \frac{6}{10} \frac{5}{9} \frac{4}{8} \frac{4}{7} + ... + \frac{4}{12} \frac{8}{11} \frac{7}{10} \frac{6}{9} \frac{5}{8} \frac{4}{7} = 6 \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot4 \cdot 4}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}=\frac{8}{33}[/tex]
For P(A=4) we can take the solution from P(A=5) and say that:
[tex]P(A=4)=c \cdot \frac{4 \cdot 3 \cdot 8 \cdot 7 \cdot 6 \cdot5 }{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}[/tex] where "c" is the combinatorial of 2 problems B with 4 problems A. In this case "c" is 15, therefore:
[tex]P(A=4)=\frac{8}{11}[/tex]
[tex]P(A \ge4)= P(A=4) + P(A=5) + P(A=6) =\frac{1}{33} +\frac{8}{33} +\frac{5}{11} = \frac{8}{11}[/tex]
