Confused about something.

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nasi112

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A stack of cards consists of six red and five blue cards. A second stack of cards consists of nine red cards. A stack is selected at random and three of its cards are drawn. If all of them are red, what is the probability that the first stack was selected?

R means three red cards
S1 means stack 1
S2 means stack 2

How to know the question is asking about P(S1 | R) or about P(R | S1)?

I assumed P(S1 | R)

[imath]\displaystyle P(S_1 | R) = \frac{P(S_1 \ AND \ R)}{P(R)}[/imath]

R = {R1,R2,R3}
S1 = {R1,R2,R3,R4,R5,R6,B1,B2,B3,B4,B5}

so the intersection S1 AND R = {R1,R2,R3} = R

But [imath]\displaystyle P(S_1 \ AND \ R) \neq P(R)[/imath] because P(R) is an independent event to both stacks. I'm confused of how I will find the probability of the intersection.
 
nasi112 said:
R = {R1,R2,R3}
S1 = {R1,R2,R3,R4,R5,R6,B1,B2,B3,B4,B5}

so the intersection S1 AND R = {R1,R2,R3} = R
Click to expand...
No, these are sets of cards, not events.

The events are
  • S1 = (randomly) choose stack 1
  • S2 = (randomly) choose stack 2
  • R = pick three cards from selected stack, and all are red
So
  • P(S1) = 1/2
  • P(S2) = 1/2
You need to calculate, for your numerator, P(S1 and R), the probability that you chose stack 1 and then all three cards you picked are red, and for the denominator, P(R) = P(S1 and R) + P(S2 and R), the probability that you either chose stack 1 and three red cards, or chose stack 2 and three red cards.

(The site kept getting an error when I tried to add this to my answer, or send it separately, until now.)
 
Dr.Peterson said:
No, these are sets of cards, not events.

The events are
  • S1 = (randomly) choose stack 1
  • S2 = (randomly) choose stack 2
  • R = pick three cards from selected stack, and all are red
So
  • P(S1) = 1/2
  • P(S2) = 1/2
You need to calculate, for your numerator, P(S1 and R), the probability that you chose stack 1 and then all three cards you picked are red, and for the denominator, P(R) = P(S1 and R) + P(S2 and R), the probability that you either chose stack 1 and three red cards, or chose stack 2 and three red cards.

(The site kept getting an error when I tried to add this to my answer, or send it separately, until now.)
Click to expand...
Thanks doctor.

This question is similar to the previous one. I don't understand why we are now suddenly restricted by events. We didn't say anything about events in the last one. I'm trying to link them together so that I can use the same idea to solve similar questions.
 
I suppose you're referring to this problem:

In a small lake, it is estimated that there are approximately 105 fish, of which 40 are trout and 65 are carp. A fisherman caught eight fish; what is the probability that exactly two of them are trout if we know that at least three of them are not?​

The current question is

A stack of cards consists of six red and five blue cards. A second stack of cards consists of nine red cards. A stack is selected at random and three of its cards are drawn. If all of them are red, what is the probability that the first stack was selected?​

The older question doesn't restrict the conditions, while this one does. Is it not obvious that the two questions are different, and therefore call for different methods??
 
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