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I'm having trouble with one of the rules of probability

P(A n B) = P(A)P(B) which holds if events A and B are independent

The following problem illustrates my confusion. I've defined Events A and B below, are these events dependent? Per the solution in the book P(A [tex]\cap[/tex] B) = P(A)P(B), that is the only way that P(A|B) will equal P(A) therefore by the solution A and B are independent

But how are they independent if the probability of selecting the next 3 cards,

P(A)= ([tex]\stackrel{11}{3}[/tex] ) / ([tex]\stackrel{50}{3}[/tex])

depends on knowing there are now only 50 cards to choose from

Moreover the solutions manual says this problem is like P(S3 S4 S5 |S1 S2) but if this is the case I see no intersection between S3 S4 S5 and S1 S2, therefore P(S3 S4 S5 [tex]\cap[/tex] S1 S2) = 0 and P(S3 S4 S5 [tex]\cap[/tex] S1 S2) != P(S3 S4 S5) P(S1 S2)

So

1) I'm not sure how these events are independant, which is the assumption taken to arrive at the answer and if they are dependent then we can't equate P(A intersect B)/P(B) to P(A)P(B)/P(B) in the solution

2) I'm a bit confused about the intersection of A and B, how is the intersection not disjoint?

Am I making a wrong assumption that A and B have nothing in common?

HERE IS THE PROBLEM:

Cards are dealt, one at a time, from a standard 52-card deck.

If the first 2 cards are both spades, what is the probability that the next 3 cards are also spades?

we break this problem down into two events

A: remaining 3 cards are spade

B: first 2 cards are spade

P(A)= ([tex]\stackrel{11}{3}[/tex] ) / ([tex]\stackrel{50}{3}[/tex])

P(B)= ([tex]\stackrel{13}{2}[/tex] ) / ( [tex]\stackrel{52}{2}[/tex] )

P(A|B) = P(A [tex]\cap[/tex] B)/P(B)

According to the book I'm working with, the answer is

P(A|B) = P(A [tex]\cap[/tex] B)/P(B)= P(A)P(B)/P(B) = P(A) = .008418

P(A n B) = P(A)P(B) which holds if events A and B are independent

The following problem illustrates my confusion. I've defined Events A and B below, are these events dependent? Per the solution in the book P(A [tex]\cap[/tex] B) = P(A)P(B), that is the only way that P(A|B) will equal P(A) therefore by the solution A and B are independent

But how are they independent if the probability of selecting the next 3 cards,

P(A)= ([tex]\stackrel{11}{3}[/tex] ) / ([tex]\stackrel{50}{3}[/tex])

depends on knowing there are now only 50 cards to choose from

Moreover the solutions manual says this problem is like P(S3 S4 S5 |S1 S2) but if this is the case I see no intersection between S3 S4 S5 and S1 S2, therefore P(S3 S4 S5 [tex]\cap[/tex] S1 S2) = 0 and P(S3 S4 S5 [tex]\cap[/tex] S1 S2) != P(S3 S4 S5) P(S1 S2)

So

1) I'm not sure how these events are independant, which is the assumption taken to arrive at the answer and if they are dependent then we can't equate P(A intersect B)/P(B) to P(A)P(B)/P(B) in the solution

2) I'm a bit confused about the intersection of A and B, how is the intersection not disjoint?

Am I making a wrong assumption that A and B have nothing in common?

HERE IS THE PROBLEM:

Cards are dealt, one at a time, from a standard 52-card deck.

If the first 2 cards are both spades, what is the probability that the next 3 cards are also spades?

we break this problem down into two events

A: remaining 3 cards are spade

B: first 2 cards are spade

P(A)= ([tex]\stackrel{11}{3}[/tex] ) / ([tex]\stackrel{50}{3}[/tex])

P(B)= ([tex]\stackrel{13}{2}[/tex] ) / ( [tex]\stackrel{52}{2}[/tex] )

P(A|B) = P(A [tex]\cap[/tex] B)/P(B)

According to the book I'm working with, the answer is

P(A|B) = P(A [tex]\cap[/tex] B)/P(B)= P(A)P(B)/P(B) = P(A) = .008418

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