To find the probability that the committee will include at least two teachers, we need to consider the different scenarios in which this can occur.
First, we will consider the scenario where exactly two teachers are selected. There are $\binom{5}{2} = 10$ ways to choose 2 teachers from the 5 available teachers, and $\binom{9}{4} = 126$ ways to choose 4 students from the 9 available students. Therefore, there are $10 \cdot 126 = 1260$ ways to form a committee with exactly two teachers.
Next, we will consider the scenario where exactly three teachers are selected. There are $\binom{5}{3} = 10$ ways to choose 3 teachers from the 5 available teachers, and $\binom{9}{3} = 84$ ways to choose 3 students from the 9 available students. Therefore, there are $10 \cdot 84 = 840$ ways to form a committee with exactly three teachers.
Finally, we will consider the scenario where all four teachers are selected (since it is not specified that all five teachers have to be selected). There is only 1 way to choose all 5 teachers, and $\binom{9}{2} = 36$ ways to choose 2 students from the 9 available students. Therefore, there are $1 \cdot 36 = 36$ ways to form a committee with all four teachers.
In total, there are $1260 + 840 + 36 = 2136$ ways to form a committee that includes at least two teachers.
To find the probability, we divide this by the total number of ways to form a committee, which is $\binom{14}{6} = 3003$ (since there are 14 people in total to choose from and we want to select 6).
Therefore, the probability is $\frac{2136}{3003} = \boxed{\frac{712}{1001}}$ or approximately 0.711.
A six member committee is formed to plan a Christmas formal. The members of the group are to be selected for five teachers and nine students. If the members are randomly selected, what is the probability that it will include at least two teachers..
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