Resolved: Quiz Question - Independent Events - card example

Didnt get it :(

Hey Pedro,

Thank you for reaching out!

Could you please specify which element of the explanation you found confusing? I will do my best to clarify.

Kind regards,

365 Hristina 

Do you mean that in order for the conditions to be able to apply for P(A|B) then there must be an intersection of a completely overlap?

Hey all,

Thank you for joining the discussion.

Let me build on what Atanas discussed above and approach the problem differently using conditional probabilities (a material covered in the Conditional Probability lecture).

We are interested in determining whether event D (drawing a Diamond) and event A (drawing an Ace) are dependent or independent. We therefore seek answers to the following two questions:

1. What is the probability of drawing a Diamond given that we've drawn an Ace, namely P(D|A)? Is this probability different from the probability of drawing a Diamond, namely P(D)?

2. What is the probability of drawing an Ace given that we've drawn a Diamond, namely P(A|D)? Is this probability different from the probability of drawing an Ace, namely P(A)?

Let's start with the first question.

1. What is the probability of drawing a Diamond given that I've drawn an Ace, namely P(D|A)? Is this probability different from the probability of drawing a Diamond, namely P(D)?

The probability of drawing a Diamond given that we've drawn an Ace is 1/4. The reason is that by drawing an Ace, we have one of four options - the Ace being either a Spade, a Heart, a Diamond, or a Club. Therefore, P(D|A) = 1/4.

Now, since there are 13 Diamonds in a deck of 52 cards, the probability of drawing a Diamond is 13/52 = 1/4. We have therefore shown that P(D|A) = P(D). That is, the probabilies of drawing a Diamond from the entire deck or from a set of four Aces are the same.

2. What is the probability of drawing an Ace given that we've drawn a Diamond, namely P(A|D)? Is this probability different from the probability of drawing an Ace, namely P(A)?

The probability of drawing an Ace given that we've drawn a Diamond is 1/13. The reason is that by drawing a Diamond, we have 13 options - all cards from 2 to 10, a Jack, a Queen, a King, and an Ace. Therefore P(A|D) = 1/13.

Now, since there are 4 Aces in a deck of 52 cards, the probability of drawing an Ace is 4/52 = 1/13. We have therefore shown that P(A|D) = P(A). That is, the probabilities of drawing an Ace from the entire deck or from a set of 13 Diamonds are the same.

This concludes the proof of events A and D being independent. Let's, however, apply the formula from the Dependent and Independent Events lecture and ensure it works for events A and D. The formula for independence is:

P(A ∩ D) = P(A) * P(D)

P(A ∩ B) is the probability of drawing a card which is simultaneously a Diamond and an Ace. The probability of drawing such a card from a standard deck is 1/52.

In the discussion above, we calculated P(A)=1/13 and P(D)=1/4. Therefore P(A) * P(D) = 1/13 * 1/4 = 1/52.

Therefore, the condition P(A ∩ D) = P(A) * P(D) is satisfied and we have again proven that events A and D are independent.

I encourage you to perform the same analysis for the rest of the answers, namely

- Drawing a Queen and drawing a Jack.

- Drawing a Heart and drawing the Jack of Hearts.

- Drawing a four and drawing the Ace of Spades.

Hope this helps!

Kind regards,

365 Hristina

Hi,

I feel the clarifications by the instructors indicates that the question should be placed after conditional probability video not here. Both formulas are mentioned in "Conditional Probability" lecture.

Also, I think the definition of indepedent events mentioned in the lecture and Q1 "The theoretical probability remains uneffected by other events" fits two answers drawing a diamond and an ace and drawing a four and the ace of spades however drawing a four can't occur at the same time with drawing the ace of space so both formulas would fail

P(A ∩ B) = 0 while P(A) * P(B) = 1/13 * 1/52 and P(A|B) = 0 = P(B|A) but does not qual P(A) not P(B)

so, I feel something missing in the defintion.

Thanks

Mo 

Hey Mohammad,

Thank you for joining the discussion!

Regarding the placement of the quiz - you're right, we'll consider moving it after the Conditional Probability lecture.

Regarding your second point, let's consider the definition of independent events stated in the lecture:

The theoretical probability remains unaffected by other events.

The way this should be interpreted mathematically is through the equalities P(A|B) = P(A) and P(B|A) = P(B). That is, neither the probability of event A is affected by event B, nor the probability of event B is affected by event A.

Let's first consider the case "drawing a Diamond (let's call this event D) and drawing an Ace (let's call this event A)". As discussed in the previous comment, P(D|A) = P(D) = 1/4 and P(A|D) = P(A) = 1/13. Therefore, since neither the probability of event D is affected by event A, nor the probability of event A is affected by event D, events A and D are independent.

Let's now consider the case "drawing a four (let's call this event F) and drawing the Ace of Spades (let's call this event AS)". As you pointed out, P(F|AS) = 0 since these events can't occur simultaneously. Moreover, P(F) = 4/52 = 1/13. Therefore the event of drawing a four is indeed affected by the event of drawing an Ace of Spades. Mathematically, this is expressed through the inequality P(F|AS) ≠ P(F).

Analogously, P(AS|F) = 0 and P(AS) = 1/52. Therefore, the event of drawing the Ace of Spades is affected by the event of drawing a four and mathematically, this is expressed through the inequality P(AS|F) ≠ P(AS). This is in compliance with the result that the events "drawing a four and drawing the Ace of Spades" are dependent.

Hope this clarifies the result! Let me know if you have further questions.

Kind regards,

365 Hristina