a. There are four 5s that can be drawn, and [tex]\binom43=4[/tex] ways of drawing any three of them. There are [tex]\binom{52}3=22,100[/tex] ways of drawing any three cards from the deck. So the probability of drawing three 5s is
[tex]\dfrac{\binom43}{\binom{52}3}=\dfrac4{22,100}=\dfrac1{5525}\approx0.00018[/tex]
In case you're asked about the probability of drawing a 3 or a 5 (and NOT three 5s), then there are 8 possible cards (four each of 3 and 5) that interest you, with a probability of [tex]\frac8{52}=\frac2{13}\approx0.15[/tex] of getting drawn.
b. Similar to the second case considered in part (a), there are now 12 cards of interest with a probability [tex]\frac{12}{52}=\frac3{13}\approx0.23[/tex] of being drawn.
c. There are four 6s in the deck, and thirteen diamonds, one of which is a 6. That makes 4 + 13 - 1 = 16 cards of interest (subtract 1 because the 6 of diamonds is being double counted by the 4 and 13), hence a probability of [tex]\frac{16}{52}=\frac4{13}\approx0.31[/tex].
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Note: [tex]\binom nk[/tex] is the binomial coefficient,
[tex]\dbinom nk=\dfrac{n!}{k!(n-k)!}={}_nC_k=C(n,k)=n\text{ choose }k[/tex]
