Help with Conditional Probability

[IVCab]

So I have this problem, I did it like this,

In a warehouse there are 10 radios of which 4 are defective, 7 televisions of which 3 are defective. 3 items are selected and they are found to be defective. What is the probability that 1 is a radio and 2 are televisions?

A = three defective items are selected.
B = 1 radio and 2 televisions are selected

They ask for: [MATH]P(B|A)=\dfrac{P(A\cap B)}{P(A)}[/MATH]
The probability of A is the probability of selecting 3 defective objects out of 17 of which 7 are defective.

P(A)= [MATH]P(A)=\dfrac{\displaystyle\binom{7}{3}}{\displaystyle\binom{17}{3}} [/MATH]
The [MATH]P(A\cap B)[/MATH] is the probability of choosing two defective televisions and one defective radio:

[MATH] P(A\cap B)=\dfrac{\displaystyle\binom{10}{1}\binom{7}{2}}{\displaystyle\binom{17}{3}} [/MATH]
Is it ok, can and I do it different?

 

[Dr.Peterson]

So I have this problem, I did it like this,

In a warehouse there are 10 radios of which 4 are defective, 7 televisions of which 3 are defective. 3 items are selected and they are found to be defective. What is the probability that 1 is a radio and 2 are televisions?

A = three defective items are selected.
B = 1 radio and 2 televisions are selected

They ask for: [MATH]P(B|A)=\dfrac{P(A\cap B)}{P(A)}[/MATH]
The probability of A is the probability of selecting 3 defective objects out of 17 of which 7 are defective.

P(A)= [MATH]P(A)=\dfrac{\displaystyle\binom{7}{3}}{\displaystyle\binom{17}{3}} [/MATH]
The [MATH]P(A\cap B)[/MATH] is the probability of choosing two defective televisions and one defective radio:

[MATH] P(A\cap B)=\dfrac{\displaystyle\binom{10}{1}\binom{7}{2}}{\displaystyle\binom{17}{3}} [/MATH]
Is it ok, can and I do it different?

Check your [MATH]P(A\cap B)[/MATH]. I think you just found [MATH]P(B)[/MATH] there.

But your strategy is a good one.

 

[Steven G]

You also never answered the problem! What does p(B|A) equal. Be careful of that error as on an exam you can do all the work correctly and then not answer the question. Worst yet you could be at a job where the boss could not care less how you got the answer, but just wanted the answer and you did not supply it.

Also why do you think that p(A U B) = p(B), is A contained in B?

I would use the reduced sample size. You are given that 3 of the chosen ones are defected. So you are choosing the 3 from from the 10 defected ones. Now just using the information on the 10 defective units what is the probability that you chose 1 radio and 2 televisions?

 

[IVCab]

I'm pretty much stucked right now, can't find [MATH]P(A\cap B) [/MATH], because @Dr.Peterson is right, I found [MATH]P(B)[/MATH] instead, I have been reading for a couple of hours but I can't really formulate the idea, if you guys could give me a little help or similar examples I can study I would aprecciate it, and @Jomo you're completly right, I didn't put the answer because I'm not that sure what I'm doing ?, but I'm tryng my best

 

[pka]

In a warehouse there are 10 radios of which 4 are defective, 7 televisions of which 3 are defective. 3 items are selected and they are found to be defective. What is the probability that 1 is a radio and 2 are televisions?

Frankly, I see nothing conditional about the question.
What is the probability of selecting one defect TV and two defective radios?
This is a simple counting question: \(\dfrac{\dbinom{4}{1}\cdot \dbinom{3}{2}}{\dbinom{17}{3}}\)

 

[IVCab]

So I've been analizing it again and I tried to finish it like this:

Remember

\(\displaystyle P(B|A)=\dfrac{P(A\cap B)}{P(A)}\) and \(\displaystyle P(A)=\dfrac{\displaystyle\binom{7}{3}}{\displaystyle\binom{17}{3}}={\dfrac{35}{680}}\)

\(\displaystyle P(A\cap B)=\dfrac{\displaystyle\binom{4}{1}\binom{3}{2}}{\displaystyle\binom{17}{3}}={\dfrac{12}{680}}\)

I changed it because of the comment @pka did,

\(\displaystyle P(B|A)=\dfrac{\dfrac{12}{680}}{\dfrac{35}{680}}=\dfrac{12}{35}=.3428\)

What do you think guys? is it ok?

 

[Dr.Peterson]

You defined
A = three defective items are selected.
B = 1 radio and 2 televisions are selected
So A∩B = 1 defective radio and 2 defective televisions are selected; that is, 1 of 4 defective radios and 2 of 3 defective televisions. That's what you correctly found the probability of.

So your answer looks good.

More simply, it turns out to be

[MATH]P(A | B)=\dfrac{\displaystyle\binom{4}{1}\binom{3}{2}}{\displaystyle\binom{7}{3}}={\dfrac{12}{35}}[/MATH]​

 

[IVCab]

Thanks guys, you have helped me a lot, now I understand most of it

 

[Steven G]

I'm pretty much stucked right now, can't find [MATH]P(A\cap B) [/MATH], because @Dr.Peterson is right, I found [MATH]P(B)[/MATH] instead, I have been reading for a couple of hours but I can't really formulate the idea, if you guys could give me a little help or similar examples I can study I would aprecciate it, and @Jomo you're completly right, I didn't put the answer because I'm not that sure what I'm doing ?, but I'm tryng my best

Here is a suggestion for an exam. If a student makes a mistake say in part a and using that wrong result in part b they make no mistake I know that I would give them full credit for part b. So on an exam if you know that you did a part wrong I advise you to continue so the grader knows that you know how to finish the problem.

 

[HallsofIvy]

Since we are given that the items selected are defective, the number of non-defective items is irrelevant. There are a total of 7 defective items, 4 radios and 3 televisions. The probability the first defective item chosen is a radio is 4/7. There are then 3 defective radios and 3 defective televisions. The probability the second defective item chosen is a television is 3/6= 1/2. There are then three defective radios and two defective televisions. The probability the third defective item chosen is a television is 2/5. The probability the three defective items chosen are a radio, and then two televisions, in that order, is (4/7)(1/2)(2/5)= 4/35.

We can show, by the same argument, that the probability the three defective items are one radio and two televisions, in any given order, is the same, 4/35. There are 3 distinct orders, RTT, TRT, and TTR, so the probability of one defective radio and two defective televisions is 3(4/35)= 12/35.