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A card is drawn from a well-shuffled pack of 52 cards. Consider two events A and B as A: a club 6 card is drawn. B: an ace card 18 drawn. Determine whether the events A and B are independent or not. - Mathematics and Statistics

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Sum

A card is drawn from a well-shuffled pack of 52 cards. Consider two events A and B as
A: a club 6 card is drawn.
B: an ace card 18 drawn.
Determine whether the events A and B are independent or not.

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Solution

One card can be drawn out of 52 cards in 52C1 ways.
∴ n(S) = `""^52"C"_1`
Let A be the event that a club card is drawn
1 club card out of 13 club cards can be drawn in 13C1 ways.
∴ n(A) = `""^13"C"_1`

∴ P(A) = `("n"("A"))/("n"("S")) = (""^13"C"_1)/ (""^52"C"_1)`
Let B be the event that an ace card is drawn.
An ace card out of 4 aces can be drawn in 4C1 ways.
∴ n(B) = `""^4"C"_1`

∴ P(B) = `("n"("B"))/("n"("S")) = (""^4"C"_1)/ (""^52"C"_1)`

Since 1 card is common between A and B
∴ n(A ∩ B) = `""^1"C"_1`

∴ P(A ∩ B) =`("n"("A" ∩ "B"))/("n"("S")) = (""^1"C"_1)/(""^52"C"_1)=1/52` .......(i)

∴ P(A) × P(B) = `(""^13"C"_1)/ (""^52"C"_1)xx(""^4"C"_1)/ (""^52"C"_1)=(13xx4)/(52xx52)=1/52` ...(ii)
From (i) and (ii), we get
P(A ∩ B) = P(A) × P(B)
∴ A and B are independent events.

Concept: Multiplication Theorem on Probability
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