Compute the conditional probability

[XodoX]

Homework Statement

A bowl contains 10 chips: 6 red chips and 4 blue chips. three chips are drawn at random and without replacement. Compute the conditional probability that

a) 2 are red and one is blue; given that at least 1 red chip is among the 3 selected
b) all are red, given that at least 2 red chips are in the sample of 3 chips.

Homework Equations

a)
A= red B= blue

The Attempt at a Solution

P(AlB)= 6/10 * 5/9 * 4/8 = 0.16= 16%.

= P(AlB)= P( A\bigcapB) / P(B)

I don't know how to do it when considering the bold. What changes ? At least one red chip and at least 2 red chips? Or can I just ignore that? I did consider all 6 red chips, so I don't know why they say "given that at least 2 red chips in the sample of 3". Of course it's more than 2, because the bowl has 6 red chips.

Homework Statement

5 cards are to be drawn successively at random and without replacement from an ordinary deck of 52 playing cards. Find the conditional probability that there are at least 3 aces in the hand, given that there are at least 2 aces.

Homework Equations

The Attempt at a Solution

I don't get this. The probability to get 3 when there are only 2 in the deck of playing cards?

 

[vela]

In this kind of problem, you want to set A to the part before the "given", so A in this case is "2 red, 1 blue". You set B to the "given" part: B = "at least one red".

Now you want to use the definition of conditional probabilityP(A\vert B)=\frac{P(A\cap B)}{P(B)}P(A\cap B) is the probability that both A and B happen. In this problem, every outcome in A is in B as well: If you chose 2 red chips, then you've chosen at least one red chip. So you have A \subset B. This means that A \cap B = A. So now you haveP(A\vert B)=\frac{P(A)}{P(B)}The probabilities P(A) and P(B) are the normal ones, not the conditional ones, you've learned to calculate in the past. So just figure those out and plug them in.

Now in the second problem, can you tell us what A and B should be?

 

[XodoX]

In this kind of problem, you want to set A to the part before the "given", so A in this case is "2 red, 1 blue". You set B to the "given" part: B = "at least one red".

Now you want to use the definition of conditional probabilityP(A\vert B)=\frac{P(A\cap B)}{P(B)}P(A\cap B) is the probability that both A and B happen. In this problem, every outcome in A is in B as well: If you chose 2 red chips, then you've chosen at least one red chip. So you have A \subset B. This means that A \cap B = A. So now you haveP(A\vert B)=\frac{P(A)}{P(B)}The probabilities P(A) and P(B) are the normal ones, not the conditional ones, you've learned to calculate in the past. So just figure those out and plug them in.

Now in the second problem, can you tell us what A and B should be?

Don't understand it. Sorry.

 

[vela]

Given you responded so quickly, I can't imagine you put much effort into understanding what I wrote.

I suggest you go back to the material from the very beginning of the course and review the concepts of outcomes and events. You really need to understand those ideas to get a good grasp of probability.

 

[XodoX]

Given you responded so quickly, I can't imagine you put much effort into understanding what I wrote.

I suggest you go back to the material from the very beginning of the course and review the concepts of outcomes and events. You really need to understand those ideas to get a good grasp of probability.

Well, you tried. Just need to explain things better.

 

[Ray Vickson]

Well, you tried. Just need to explain things better.

No, you need put in more effort. Do you have a textbook, and have you read it? Do have course notes and have you consulted them? Have you searched on-line for related concepts?

RGV

 

[XodoX]

No, you need put in more effort. Do you have a textbook, and have you read it? Do have course notes and have you consulted them? Have you searched on-line for related concepts?

RGV

I can get the same from a textbook. That's no help.

 

[Redbelly98]

You have to do some "counting" calculations. How many different ways can you get 2 red chips and 1 blue? How many different ways can you get at least 1 red chip?