Probability of Buying Faulty Brand A TV

[andrey21]

In a shop 1/3 of the TV's are brand A, the probability that a brand A TV is faulty is 0.2. What is the probability person 1 buys a faulty brand A TV?
Here's my attempt
Im not sure if it seems almost too easy but should I use the rule:

P(A) x P(B) = P(AnB)

Giving:

0.33333x0.2=0.066666 ??

 

[susskind_leon]

That's pretty much it

 

[andrey21]

Thanks Susskind, think I convinced myself it was too simple

 

[andrey21]

Just a quick follow up, what would be the probability for the 11th customer? Would it just be a case of multiplying 0.066666 by 20?

 

[susskind_leon]

Can you specify for question a little bit? The probability that the 11th customer buys a faulty brand A TV?

 

[andrey21]

what is the expected number of faulty TV's after 11 customers?

 

[andrey21]

So since the probability of a faulty TV for person 1 is 0.06666 or 1/15, after 11 customers the probability would be:

11 x 1/15 =0.733333

Since we can't have 0.7333 of a TV I just round this to 1. Therefore 1 faulty TV after 11 customers.

 

[Ray Vickson]

Just a quick follow up, what would be the probability for the 11th customer? Would it just be a case of multiplying 0.066666 by 20?

I seem to be missing something: your initial post asked about the probability that person 1 gets a faulty Brand A TV, but mentioned nothing about how many TVs the store has in stock, or how many customers buy TVs. Was this info posted in another message that, for some reason, my computer cannot access? Could you please repeat the context, as there may be some serious and tricky issues involved.

RGV

 

[andrey21]

Basically I am asked to establish the probability that customer 1 will get a faulty brand A TV. Given that there are 1/3 of TV's in shop are A and probability of faulty TV is 0.2.

Hence: 1/3 x 1/5 = 1/15 or 0.06666666

Now the question switches to ask how many faulty brand A TV's will be purchased if 11 customers brought brand A TV's?

Since I know there is a 1/15 chance of a brand A TV being faulty, then would this lead to 11x1/15 = 0.73333

 

[Ray Vickson]

Basically I am asked to establish the probability that customer 1 will get a faulty brand A TV. Given that there are 1/3 of TV's in shop are A and probability of faulty TV is 0.2.

Hence: 1/3 x 1/5 = 1/15 or 0.06666666

Now the question switches to ask how many faulty brand A TV's will be purchased if 11 customers brought brand A TV's?

Since I know there is a 1/15 chance of a brand A TV being faulty, then would this lead to 11x1/15 = 0.73333

Yes, I got the first part; it was the second part I wanted to see. Now you need to be careful: you said *11 customers bought brand A*, so the "1/3" is no longer relevant: now you look only at P{brand A is faulty} = 0.20 for each of the 11 sold. So the number faulty sold is a binomial random variable with n = 11 and p = 0.20. The expected number faulty type A sold is n*p = 2.2.

Had you said 11 people bought TV's, the number of those that are faulty of type A would be binomial with parameters n = 11 and p = 1/15; but that is not what you said.

RGV

 

[andrey21]

For the second answer you gave would that figure be obviously 11 x 1/15 = 0.733333 faulty type A TV's. So that would be 1 in every 11 customers gets a faulty brand A TV.

 

[Ray Vickson]

I would never say that because it its not true. The EXPECTED number faulty is not 1 (in your 11/15 case); it is 0.7333... . The (actual) number getting a a faulty TV is a _random variable_ that can have values 0, 1, 2, ..., 11, with probabilities given by the binomial distribution. You really do need to distinguish between "number" and "expected number".

RGV

 

[andrey21]

Ok so the expected number then would that be 0.733333? But the actual number would of course be one because we cannot have 0.7333 of a faulty TV

 

[Ray Vickson]

No. The actual number would be 0 or 1 or 2 or 3 or ... . In fact, the most probable number is 0 and the median number is 0 in this case.

Please forget about trying to interpret "expected" values in terms of ordinary, everyday usage of the word "expected". In probability and statistics the word "expected" has a definite, technical meaning that may not (and often does not) match our normal usage. Same word, different meanings. This is really the same situation as, for example, saying "I had an argument with my girlfriend" vs. asking about the argument of a sine function vs. taking the argument of a complex number. These are three different uses of the word 'argument', and it is the same with the word 'expected' or 'expectation'.

RGV

 

[andrey21]

Ok Thankyou Ray Vickson just to confirm though the expected number I posted (0.733333) that is correct?

 

[Ray Vickson]

Ok Thankyou Ray Vickson just to confirm though the expected number I posted (0.733333) that is correct?

Yes.

RGV