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## Comment on

Determining Independence## What is the answer for the 2

## P(both selections are women)

P(both selections are women) = P(1st selection is a woman AND 2nd selection is a woman)

= (4/7)(3/6)

= 12/42

= 2/7

## what is the answer for both

## P(1st toss is heads AND 2nd

P(1st toss is heads AND 2nd toss is heads)

= P(1st toss is heads) x P(2nd toss is heads)

= 1/2 x 1/2

= 1/4

## In the two women selection

what's wrong with this method

prob = 3c2/7c2

order doesn't matter so why can't we use this method

## Your solution is great,

Your solution is great, EXCEPT for the numerator (it's not 3C2).

P(both selections are women) = (number of ways to select 2 women from the FOUR women)/(number of ways to select 2 people from the 7 people)

= (4C2)/(7C2)

= 6/21

= 2/7

## oh my bad Thanks a lot :)

Thanks a lot :)

## in the women case, even a

## That's true!

That's true!

## Can you help me understand

https://www.khanacademy.org/math/statistics-probability/probability-library/conditional-probability-independence/e/identifying-dependent-and-independent-events

## Here's the question: "Suppose

Here's the question: "Suppose that Adam rolls a fair six-sided die and a fair four-sided die simultaneously. Let A be the event that the six-sided die is an even number, and let B be the event that the four-sided die is an odd number."

Notice that the probability of event B (getting an odd number on the 4-sided die) is NOT affected by whether or not event A occurs.

Regardless of what happens with the 4-sided die, P(B) is ALWAYS 1/2

Compare this to the following question:

A bag contains 5 red balls and 1 yellow ball.

Joe randomly select one ball (without replacement), and then another ball.

Let event A be selecting the yellow ball on the first draw.

Let event B be selecting the yellow ball on the second draw.

What is the probability of event B?

Well, it DEPENDS.

If event A (getting a yellow ball on the first draw) occurs, then P(event B) = 0, since the yellow ball has already been taken.

If event A does NOT occur, then P(event B) = 1/5, since there are now 5 balls remaining, and 1 of them is yellow.

As you can see, the probability of event B occurring DEPENDS on whether event A occurred.

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