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I am totally new to probability and I am a little bit confused. I have the following homework:

A large group of people are competing for all-expense-paid weekends in Philadelphia. The Master of Ceremonies gives each contestant a well-shuffled deck od cards. The contestant deals two cards off the top of the deck, and wins a weekend if the first card is the ace of hearts or the second card is the king of hearts. What is the probability of wining the weekend?

I tried to solve this exercise in three ways:

  1. Using $P(A ∪ B) = P(A) + P(B) - P(A ∩ B)$. I get:$$\frac{1}{52} + \frac{1}{52} - \frac{1}{52}×\frac{1}{51} = \frac{101}{51×52}.$$
  2. Using $P(A ∪ B) = P(A) + P(B ∩ A^c)$. I get:$$\frac{1}{52} + \frac{1}{52}×\frac{50}{51} = \frac{101}{51×52}.$$
  3. Using formula $P(A) = 1 - P(A^c)$ where the opposite is not getting the ace of hearts as the first card and not getting the king of hearts as the second card. In this way I get:$$1 - \frac{51}{52}×\frac{50}{51} = \frac{2}{52} \ne \frac{101}{51×52}.$$

What am I doing wrong in the third way? Thank you in advance for your help.

Ѕᴀᴀᴅ
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Steve Cheng
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2 Answers2

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The problem with your computation in #3 is that you didn't account for the possibility that the first card was the king of hearts.

Michael Biro
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  • Thank you very much, mister. I think I understand my mistake but I still don't see how to write the valid formula in this case. Or is there even any sense in solving this problem in this way? – Steve Cheng Mar 26 '19 at 02:04
  • Sure you could, one way is just to break it into cases. For example $1 - (\frac{50}{52}\cdot\frac{50}{51} + \frac{1}{52}\frac{51}{51})$ – Michael Biro Mar 26 '19 at 02:11
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    Just modify the symbols a bit. Let $A_k$ be the event of drawing an ace of hearts on draw $k$, and $B_k$ be the event of drawing king of hearts on draw $k$.$$\mathsf P(A_1\cup B_2)~{=1-\mathsf P(A_1^\complement\cap B_2^\complement)\=1-\mathsf P(A_1^\complement\cap B_1\cap B_2^\complement)-\mathsf P(A_1^\complement\cap B_1^\complement\cap B_2^\complement)\=1-\mathsf P(B_1)-\mathsf P((A_1\cup B_1)^\complement\cap B_2^\complement)\=1-\tfrac{1}{52}-\tfrac{50}{52}\tfrac{50}{51} }$$ – Graham Kemp Mar 26 '19 at 02:12
  • Now I get it but I must admit that I needed some time to wrap my mind around the correct solution and to understand the nuance introduced by the possibility of having the king of hearts as the first card. Thank you both, your answers were very helpful. – Steve Cheng Mar 26 '19 at 02:56
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In the third way, the formula is ($A-$ ace of hearts, $B-$ king of hearts): $$P(A\cup B)=1-P(A^C\cap B^C)=1-\frac{51}{52}\cdot \color{red}{\frac{50}{51}},$$ however, the event $B^C$ depends on the event $A^C$. In other words, you assumed the first card is not king of hearts, however: $$P(B_2^C|B_1)=1; P(B_2^C|B_1^C)=\frac{1}{51}.$$

Probability tree diagram ($A'=A^C$):

$\hspace{1cm}$enter image description here

$$B\cup B'=S; A'\cap S=A';\\ P(A'\cap B')=P(\color{blue}{(A'\cap (B\cup B'))}\cap \color{green}{B'})=\\ P(\color{blue}{([A'\cap B]\cup [A'\cap B'])}\cap \color{green}{B'})=\\ P(\{\color{blue}{[A'\cap B]}\cap \color{green}{B'}\}\cup \{\color{blue}{[A'\cap B'])}\cap \color{green}{B'}\})=\\ P(\color{blue}{[A'\cap B]}\cap \color{green}{B'})+P(\color{blue}{[A'\cap B'])}\cap \color{green}{B'})=\\ \frac{1}{52}\cdot \frac{51}{51}+\frac{50}{52}\cdot \frac{50}{51}.$$ Note: The events in blue color are the first card, while in green color are the second card. They can be differentiated by relevant indices (subscripts) as labeled by Graham Kemp in his comment.

Hence: $$\begin{align}P(A\cup B)&=1-P(A'\cap B')=\\ &=1-\left(\frac{1}{52}\cdot \frac{51}{51}+\frac{50}{52}\cdot \frac{50}{51}\right)=\\ &=\frac{101}{51\cdot 52}.\end{align}$$

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