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Mathematics

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malfutka [58]2 years ago

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a. There are four 5s that can be drawn, and $\binom43=4$ ways of drawing any three of them. There are $\binom{52}3=22,100$ ways of drawing any three cards from the deck. So the probability of drawing three 5s is

$\dfrac{\binom43}{\binom{52}3}=\dfrac4{22,100}=\dfrac1{5525}\approx0.00018$

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 $\frac8{52}=\frac2{13}\approx0.15$ of getting drawn.

b. Similar to the second case considered in part (a), there are now 12 cards of interest with a probability $\frac{12}{52}=\frac3{13}\approx0.23$ 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 $\frac{16}{52}=\frac4{13}\approx0.31$ .