07 Mar 2024

Posted on:

12 May 2021

3

# Resolved:Quiz Question - Independent Events - card example

Hi,

I find it difficult to understand why the correct answer to the question about independent sets is:
"Drawing a Diamond and drawing an Ace.". I hope you can help me understand where I am going wrong with this.

As I understand it, the probabilities for both events must be equal, P(A) = P(B), to say that they are independent. Meanwhile, when we draw a Diamond, it influences the probability of drawing an Ace, since we might draw an Ace of Diamonds, right? What am I missing?

Instructor
Posted on:

08 Oct 2021

10

Hello Hubert,

Strictly speaking, two events A and B are independent when the following formula holds: P(AB) = P(A) * P(B).
This here is the definition of when two events A and B are independent (and not necessarily that they have equal probability). Being independent means not being influenced by the other event occuring, so it is a good idea to always think of the events taking place at the same time.

In the question, let A be 'drawing a Diamond' and B be 'drawing an Ace'. Calculate the probabilities:
P(AB) - this is the probability of drawing one card, which is simultaneously a Diamond and an Ace. Because there is only one such card in the deck (namely, the Ace of Diamonds), the probability of that occuring is exactly 1 /52. On the other hand, knowing that P(A) = 1 / 4 and P(B) = 1 / 13, we verify that indeed events A and B are independent.

Please consider the other examples in a similar manner and try to ponder why the events are not independent.
Hint - let's see what happens if A is 'drawing a four' and B is 'drawing the Ace of Spades'. Now, you get P(A) = 1 / 13 and P(B) = 1 / 52, but P(AB) = 0 because you cannot possibly draw one card which is simultaneously a four and the Ace of Spades. Try the other examples on your own.

Hope this helps!

Best,
A. The 365 Team

Posted on:

12 Sept 2023

0

Didnt get it :(

Instructor
Posted on:

12 Sept 2023

0

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

Posted on:

12 Sept 2023

0

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?

Instructor
Posted on:

27 Sept 2023

0

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)?

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.

Instructor
Posted on:

27 Sept 2023

0

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

Posted on:

06 Mar 2024

0

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

Instructor
Posted on:

07 Mar 2024

2

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