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 to the probability of an answer of the X group to be the i° picked problem from the j° picked problem of that group:
with the total number of problems in that group.
We analyze now 3 different problems:
For P(A=4) we can take the solution from P(A=5) and say that:
where "c" is the combinatorial of 2 problems B with 4 problems A. In this case "c" is 15, therefore:
Answer:
Step-by-step explanation:
Ah, probablity.
Since their are 2 sides of a coin, it would be for both.
×
Answer:
1/4a -1/6b + 1/10c
Step-by-step explanation:
1/2 a -1/3 b + 1/5c + -1/4a + 1/6 b - 1/10c
Combine like terms
1/2a - 1/4a -1/3b + 1/6b +1/5c - 1/10 c
2/4a -1/4a -2/6b + 1/6b +2/10c -1/10c
1/4a -1/6b + 1/10c
Answer:
Step-by-step explanation:
Your answer will be 1:9
Answer:
say less
Step-by-step explanation:
Give me a mc flurry
:-)