Explanations:The problem involves calculating the probability of selecting at least one student from Karnataka when four students are chosen at random from a class of 15 students, where 7 are from Maharashtra, 5 from Karnataka, and 3 from Goa. To solve this, we use the concept of complementary events. The complementary event to "at least one student from Karnataka" is "no students from Karnataka."
First, we calculate the total number of ways to select 4 students out of 15. This is given by the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of items, and \( r \) is the number of items to choose. Here, \( n = 15 \) and \( r = 4 \):
\[
\binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15!}{4! \cdot 11!} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365
\]
Next, we calculate the number of ways to select 4 students from the 10 students who are not from Karnataka (7 from Maharashtra and 3 from Goa). Here, \( n = 10 \) and \( r = 4 \):
\[
\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4! \cdot 6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210
\]
The probability of selecting no students from Karnataka is the ratio of the number of ways to select 4 students from the 10 non-Karnataka students to the total number of ways to select 4 students from 15:
\[
P(\text{no students from Karnataka}) = \frac{\binom{10}{4}}{\binom{15}{4}} = \frac{210}{1365} = \frac{2}{13}
\]
The probability of selecting at least one student from Karnataka is the complement of the above probability:
\[
P(\text{at least one student from Karnataka}) = 1 - P(\text{no students from Karnataka}) = 1 - \frac{2}{13} = \frac{13}{13} - \frac{2}{13} = \frac{11}{13}
\]
Thus, the probability that at least one student is from Karnataka is \( \frac{11}{13} \). This approach leverages the complementary event to simplify the calculation, making it easier to find the desired probability.