## Related questions with answers

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.

Solution

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

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

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

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

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

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

Definition Conditional probability:

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

Use the definition of conditional probability:

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

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