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Independence of events is not always obvious. Toss two balanced coins independently. The four possible combinations of heads and tails in order each have probability 0.25. The events A = head on the first toss B = both tosses have the same outcome may seem intuitively related. Show that P(B|A) = P(B), so that A and B are in fact independent.

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We have 1 chance in 2 to toss heads with a fair or balanced coin:

P(A)=P( head on the first toss)=12=0.5P(A)=P(\text{ head on the first toss})=\dfrac{1}{2}=0.5

The four possible combinations of heads and tails in order each have probability 0.25 (given in exercise prompt):

P(B)=P( both tosses have the same outcome )=P(2 heads)+P(2 tails)=0.25+0.25=0.50P(B)=P(\text{ both tosses have the same outcome })=P(2\text{ heads})+P(2\text{ tails})=0.25+0.25=0.50

If we have a head on the first toss and both tosses have the same outcome, then both tosses have heads:

P(A and B)=P( heads)=0.25P(A\text{ and }B)=P(\text{ heads})=0.25

Definition Conditional probability:

P(BA)=P(A and B)P(A)P(B|A)=\dfrac{P(A\text{ and }B)}{P(A)}

Use the definition of conditional probability:

P(BA)=P(A and B)P(A)=0.250.5=12=0.50=P(B)P(B|A)=\dfrac{P(A\text{ and }B)}{P(A)}=\dfrac{0.25}{0.5}=\dfrac{1}{2}=0.50=P(B)

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