09 Jan 2024

Posted on:

12 Sept 2023

1

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

Hi,

I was at the question 2/3 of quiz for Dependent and Independent Events, and I just have my doubt about the correction answer...wouldn't it be drawing a Queen and drawing a Jack the independent events instead of drawing a Diamond and drawing an Ace?

Instructor
Posted on:

12 Sept 2023

0

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

Posted on:

26 Sept 2023

1

Instructor
Posted on:

27 Sept 2023

2

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)?

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

2

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:

09 Jan 2024

1

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