Apologies for the formatting, I am not familiar with this platform.
I believe I have solved it but I looking for assurance.
Let X be a random variable indicating the number of cards of the same suit (Spade, Diamond, Club, Heart)
I solved the problem by looking at the probability of drawing 5 cards of a particular suit and multiplying that number by 4.
Total number of possibilities when drawing seven cards = nCr(52,7) = 133784560 = T
P(X≥5) = P(X=5) + P(X=6) + P(X=7)
P(X≥5) = [nCr(13,5)/T] + [nCr(13,6)/T] + [nCr(13,7)/T]
P(X≥5) = 3.527*10^-5 = probability of getting 5 cards of a particular suit
Then I multiplied the answer by 4 to obtain the probability of getting 5 cards of any suit
(3.52710^-5) * 4 = 1.4110^-4 = 0.0141%
What do you think?
The probability of exactly five spades has $\binom{13}{5}$ in the numerator, yes, for choosing which five spades those were. It also has an additional $\binom{39}{2}$ in the numerator for choosing which non-spades were also included in the hand. Your attempt is missing these additional terms in your numerators. See the Hypergeometric Distribution.
Apart from this, your attempt was correct.
The correct probability is $$4\cdot \dfrac{\binom{13}{5}\binom{39}{2}+\binom{13}{6}\cdot 39 + \binom{13}{7}}{\binom{52}{7}}\approx 0.03056577$$