Why drawing a queen and drawing a jack is NOT an independent event?

Hey Sherine,

Thank you for reaching out!

Please refer to the following thread where a similar question is resolved:

https://365datascience.com/q/782afba348

Kind regards,

365 Hristina

I have read the thread, but didnot understand. Can you please explain in detail

Hey Fahad,

Thank you for reaching out.

Let's approach the problem differently and use conditional probabilities (using the material from the lecture following the quiz).

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

Let's see option wise

Option A: Drawing a queen and drawing  Jack

Not feasible. If you know, you have drawn a queen, how can the card be  Jack?

If you have drawn a queen the probability of drawing a jack becomes 0. We have to think that we are drawing a single card.

Option B: Drawing a heart and Drawing a jack of Heart

When we draw the heart card out, prob. for drawing out Jack of Heart changes to 1/13 (initially 1/52) 

Option C:

Drawing a diamond does not change prob. of drawing ace(initially 4/52  and after drawing diamond 1/13)

Option D:

Drawing of 4 changes the prob second to 0