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Firstly, i want to sorry for asking the question that has been answered. This is because i don't have enough reputation so that i can ask the people who gave the answer.

The question that has been answered: Independent events, drawing cards without replacement.

Question: Two cards are chosen from a pack of cards without replacement. Are the following events independent? (i)the first card is a heart, (ii)the second card is a picture card.

After reading this, i still can not understand why these 2 events are independent. I can prove these 2 are independent:

$A$ : First card is a heart.
$B$ : Second card is a picture card.

\begin{align} P(A) &= \frac{13}{52} \\ P(B) &= \frac{12}{52} \end{align}
If we assume that these two events are independent then : \begin{align} P(A \cap B) &= P(A).P(B) = \frac{13.12}{52^2} = \frac{3}{52} \end{align} We also have: \begin{split} P(A \cap B) &= P(\text{first card is a heart card with picture }\cap B) \\ &\quad+ P(f\text{first card is a heart card not picture }\cap B)\\ &=\frac{3}{52}\cdot \frac{11}{51} \; + \frac{10}{52}\cdot\frac{12}{51} = \frac{3}{52} \end{split} So that: \begin{align} P(A \cap B) &= P(A)\cdot P(B)\;\;\textit{is true} \end{align} This proves that these two events are independent.

The reason here is that i think that when we pick the first card is a heart then we have two cases :

1) If we pick the first card, it is a heart picture card. So the number of picture cards and total cards in the pack of cards decreases.

2) If we pick the first card, it is a heart but not a picture card. So the number of total cards in the pack of cards decreases.

I think on both cases these 2 events are dependent because after the event $A$ happens, it affects to the probability of the event $B$.

After thinking a lot i still can not figure out why they are independent. Thanks a lot for reading and helping me !

Daniele Tampieri
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Alex
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    Possible duplicate, from the Related links to this question https://math.stackexchange.com/questions/3002068/counter-intuitive-independent-and-dependent-events?rq=1 – Ethan Bolker Apr 30 '19 at 17:47

1 Answers1

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I think the error in your intuition is here:

because after the event A happens, it affects to the probability of the event B.

That is correct only if you have been told whether the card chosen on the first pick was a face card, not just that it was a heart. Without that information you know nothing that can change the probability that the second card is a face card.

In fact you don't even care whether the first card is a heart. If you pick one card from the deck and throw it away without looking then the probability that the next card is a face card is $12/52$.

Edit in response to the comments.

Suppose there are just four cards in the deck, FXXX, and you throw one away. What is the probability that the next card is the F? Clearly it's $$ \frac{1}{4} \times 0 + \frac{3}{4} \times \frac{1}{3} = \frac{1}{4} $$ where the two cases in the sum come from picking the F first or not.

You can do the same algebra for the full problem. The answer is $12/52$, as it was from the start, not $11/51$ or $12/51$.

Ethan Bolker
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    Can you explain why "the probability that the next card is a face card is 12/52" ? Because i think that when we pick a card from the pack of cards the remain number of cards is 51 – Alex Apr 30 '19 at 17:23
  • @Alex I agree to you. – callculus42 Apr 30 '19 at 17:30
  • @callculus See my edit. The denominator is $52$, not $51$. – Ethan Bolker Apr 30 '19 at 17:42
  • Yeah, that´s true. – callculus42 Apr 30 '19 at 17:47
  • @EthanBolker Thanks a lot now i understand why "the probability that next card is a face card is 12/52". However, i still don't understand why "Without that information" the first event can not change the probability of the second event, because i think that when we draw a first card, it may be a heart face card or a heart but not face card. – Alex May 01 '19 at 03:15
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    Indeed when you draw that first heart it may be a face card or not. The two probabilities $3/13$ and $10/13$ there exactly balance the change in face probability for the second card, leaving it at $12/52$. That's essentially what the calculations in my answer and in your question show. – Ethan Bolker May 01 '19 at 12:00