This is a multi part question answered separately on Clay6.com

- If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)

In a deck there are 52 cards - 13 of each suit - spades, clubs, diamonds and hearts, and there are 4 aces, one in each suit.

Given E: the card drawn is a spade, P(E) = $\large\frac{13}{52} = \frac{1}{4}$

Given F: the card drwan is an ace, P(F) = $\large\frac{4}{52} = \frac{1}{13}$

If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)

P (spade $\cap$ ace) = P (E $\cap$ F) = P (drawing an ace of spades) = $\large\frac{1}{52}$

P (E $\cap$ F) = P(E) $\times$ P(F) = $\large\frac{1}{4}$$\;\times$$\large\frac{1}{13} = \frac{1}{52}$

Therefore E and F are independent

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