Probability Theory, work check

[WWCY]

Homework Statement

Hi all, could someone give my working a quick skim to see if it checks out? Many thanks in advance.

Suppose that 5 cards are dealt from a 52-card deck. What is the probability of drawing at least two kings given that there is at least one king?

Homework Equations

The Attempt at a Solution

Let $B$ denote the event that at least 2 kings are drawn, and $A$ the event that at least 1 king is drawn. Because $B$ is a strict subset of $A$,
$$P(B|A) = P(A \cap B)/P(A) = P(B)/P(A)$$
Compute $P(A)$, $P(A^c )$ denotes the probability of not drawing a single king.
$$P(A) = 1 - P(A^c) = 1 - \frac{48 \choose 5}{52 \choose 5} \approx 1 - 0.6588$$
Compute $P(B)$, $P(B^c)$ denotes the probability of not drawing at least 2 kings, which is the sum of probabilities of drawing 1 king $P(1)$ and the probability of not drawing a single king $P(A^c)$.
$$P(B) = 1 - P(B^c) = 1 - (P(1) - P(A^c))$$
$$P(1) = \frac{5 \times {4\choose 1} \times 48 \times 47 \times 46 \times 45 }{52 \times 51 \times 50 \times 49 \times 48} \approx 0.299$$
where the numerator is the number of ways one can have a hand of 5 containing a single king.
$$P(B) \approx 1 - (0.299 + 0.6588) \approx 0.0422$$
finally,
$$P(B|A) = P(B)/P(A) = 0.0422 / (1-0.6588) \approx 0.1237$$

I'm pretty sure that there is a quicker way to do all of this even if my work checks out, I'd appreciate if someone could demonstrate a more efficient calculation!

 

[verty]

I agree with how you have calculated it. I get 0.1222.

 

[PeroK]

I get the same answer as @verty.

The way I looked at it was:

$p = \frac{1 - p_0 - p_1}{1-p_0} = 1 - \frac{p_1}{1-p_0}$

Where $p_n$ is The probability of drawing exactly $n$ kings.

 

[Ray Vickson]

Homework Statement

Hi all, could someone give my working a quick skim to see if it checks out? Many thanks in advance.

Suppose that 5 cards are dealt from a 52-card deck. What is the probability of drawing at least two kings given that there is at least one king?

Homework Equations

The Attempt at a Solution

Let $B$ denote the event that at least 2 kings are drawn, and $A$ the event that at least 1 king is drawn. Because $B$ is a strict subset of $A$,
$$P(B|A) = P(A \cap B)/P(A) = P(B)/P(A)$$
Compute $P(A)$, $P(A^c )$ denotes the probability of not drawing a single king.
$$P(A) = 1 - P(A^c) = 1 - \frac{48 \choose 5}{52 \choose 5} \approx 1 - 0.6588$$
Compute $P(B)$, $P(B^c)$ denotes the probability of not drawing at least 2 kings, which is the sum of probabilities of drawing 1 king $P(1)$ and the probability of not drawing a single king $P(A^c)$.
$$P(B) = 1 - P(B^c) = 1 - (P(1) - P(A^c))$$
$$P(1) = \frac{5 \times {4\choose 1} \times 48 \times 47 \times 46 \times 45 }{52 \times 51 \times 50 \times 49 \times 48} \approx 0.299$$
where the numerator is the number of ways one can have a hand of 5 containing a single king.
$$P(B) \approx 1 - (0.299 + 0.6588) \approx 0.0422$$
finally,
$$P(B|A) = P(B)/P(A) = 0.0422 / (1-0.6588) \approx 0.1237$$

I'm pretty sure that there is a quicker way to do all of this even if my work checks out, I'd appreciate if someone could demonstrate a more efficient calculation!

Your method is OK. The method of Perok in #3 is faster. However, I have one quibble: you ought to keep more significant figures when doing calculations that involve subtractions, so as to avoid "subtractive error magnification". In your case you do not do too badly, getting 0.1237 instead of 2257/18472 ≈ 0.12218, but the general principle still holds. (In some problems subtractive error magnification can lead to huge errors, perhaps even incorrect final signs, etc.)

 

[WWCY]

Your method is OK. The method of Perok in #3 is faster. However, I have one quibble: you ought to keep more significant figures when doing calculations that involve subtractions, so as to avoid "subtractive error magnification". In your case you do not do too badly, getting 0.1237 instead of 2257/18472 ≈ 0.12218, but the general principle still holds. (In some problems subtractive error magnification can lead to huge errors, perhaps even incorrect final signs, etc.)

Ah okay, I'll keep that in mind the next time. Thanks!