1
$\begingroup$

We have a condensed deck of 40 cards containing only the denominations from Ace through 10. We draw four cards uniformly at random (without replacement).

  1. What is the probability that we draw four distinct denominations, such as one Ace, one five, one six and one seven.

  2. What is the probability that we draw at least three hearts.

For 1, let's say we divide the the 40 cards into 10 piles of 4 cards each (of each kind Ace to 10). Then there are 10C4 ways to choose 4 piles and 4C1 ways to choose a card from each pile, so the probability should be ((10C4)(4)(4C1))/(40C4) = 0.036. Is this correct or am I missing something?

For 2, I think something like 1 - Pr(choosing exactly 2 hearts) could work, but I don't know where to go from here.

Any help would be appreciated.

$\endgroup$
7
  • $\begingroup$ the complement of "throwing at least $3$ hearts" is not "throwing exactly $2$ hearts." $\endgroup$
    – lulu
    Apr 12 at 23:41
  • $\begingroup$ I don't understand your calculation for the first one, but your result is far too small. Try it out to get some insight. $\endgroup$
    – lulu
    Apr 12 at 23:45
  • $\begingroup$ Ah, there should be a factor of $4^4$ and you only have $4^2$. For each of the four separate ranks youy have selected, there are four choices, hence $4^4$. $\endgroup$
    – lulu
    Apr 12 at 23:46
  • $\begingroup$ Oh yeah you're right, Thank you! Also, for part2, could I do something like Pr(3H and 1 other) + Pr(4H) divided by (40C4) $\endgroup$
    – rubik
    Apr 12 at 23:51
  • $\begingroup$ Yes, that would be fine. As for the probability of that happening, you can approach as you are or you can go straight to the end result by citing the hypergeometric distribution (which is probably going to be the same as what you're doing, it is just the formal name for it) $\endgroup$
    – JMoravitz
    Apr 12 at 23:56
2
$\begingroup$

For 1. $P=\frac{36\times 32\times 28}{39\times 38\times 37}=0.588248$.

For 2. There are two terms: exactly three hearts and exactly four hearts.

$P_3=4\times\frac{30\times 10\times 9\times 8}{40\times 39\times 38\times 37}=0.0393916$ $P_4=\frac{10\times 9\times 8\times 7}{40\times 39\times 38\times 37}=0.00229784$

Total $P=0.04168944$

$\endgroup$
1
  • $\begingroup$ 2. has been corrected. $\endgroup$ Apr 13 at 2:50
1
$\begingroup$

We have a condensed deck of 40 cards containing only the denominations from Ace through 10.

Presumably that is those ten denominations in each of the four standard suits.

We draw four cards uniformly at random (without replacement).

What is the probability that we draw four distinct denominations, such as one Ace, one five, one six and one seven.

Well, we are comparing the count for ways to select four from ten denominations, with each in one from four suits, to the count for ways to select four from forty cards.   That "each" means we are not multiplying ${^{4}\mathrm C_1}$ by $4$ but rather empowering it. $$\dfrac{{^{10}\mathrm C_4}\,({^{4}\mathrm C_1})^4}{^{40}\mathrm C_4} = \dfrac{40\cdot 36\cdot 32\cdot 28}{40\cdot 39\cdot 38\cdot 37}$$

What is the probability that we draw at least three hearts.

For 2, I think something like 1 - Pr(choosing exactly 2 hearts) could work, but I don't know where to go from here.

That is not the correct complement.

The event of interest is drawing 3 or 4 from the four hearts (with the remainder of cards drawn being other suits). The complement of this is drawing 0, 1, or 2 from the four hearts.

Thus using the rule of complements is contra-indicated -- as calculating the probability for the complement would take more effort. (However, it will still get you the answer).

And you just use the rule of addition: the probability for the union of disjoint events is the sum of the probabilities for those events.   So add the probability for drawing three hearts to the probability for drawing four hearts.

$$\dfrac{{^4\mathrm C_3}\,{^{36}\mathrm C_1}+{^4\mathrm C_4}\,{^{36}\mathrm C_0}}{^{40}\mathrm C_4}=1-\dfrac{{^4\mathrm C_0}\,{^{36}\mathrm C_4}+{^4\mathrm C_1}\,{^{36}\mathrm C_3}+{^4\mathrm C_2}\,{^{36}\mathrm C_2}}{^{40}\mathrm C_4}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.