This is the question, and i dont really know where to start.. --> Suppose that a fair coin is tossed four times. 'A' is the event that the first toss yields heads and 'B' is the event that 2 heads and 2 tails occur. Are 'A' and 'B' independent events? Explain.
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Recall that P(B l A) = P(B n A) / P(A).
Now P(A) = 1/2. There are 2^4 = 16 possible outcomes from the four tosses.
8 of these outcomes have a head as the first toss, and 3 of these 8 outcomes
are such that there are a total of 2 heads and 2 tails after 4 tosses.
Thus P(B n A) = 3/16, and so P(B l A) = (3/16) / (1/2) = 3/8.
Next, P(B) = [4!/(2!*2!)] / 16 = 6/16 = 3/8. Since P(B l A) = P(B)
we conclude that B independent of A.
Now P(A l B) = P(A n B) / P(B) = (3/16) / (3/8) = 1/2 = P(A), so we can
conclude that A is independent of B.
Thus A and B are independent events.
Now P(A) = 1/2. There are 2^4 = 16 possible outcomes from the four tosses.
8 of these outcomes have a head as the first toss, and 3 of these 8 outcomes
are such that there are a total of 2 heads and 2 tails after 4 tosses.
Thus P(B n A) = 3/16, and so P(B l A) = (3/16) / (1/2) = 3/8.
Next, P(B) = [4!/(2!*2!)] / 16 = 6/16 = 3/8. Since P(B l A) = P(B)
we conclude that B independent of A.
Now P(A l B) = P(A n B) / P(B) = (3/16) / (3/8) = 1/2 = P(A), so we can
conclude that A is independent of B.
Thus A and B are independent events.