# A coin is tossed successively three times. Find the probability of getting exactly one head or two heads.

## Solution :

We know that probability of event = no. of favorable outcome/total outcome

Sample space of event : { (H , H , H) , (H , H , T) , (H , T , H) , (H , T , T) , (T , H , H) , (T , H , T) , (T , T , H) , (T , T , T) }

favorable outcome of getting one head : { (H , T , T) , (T , H , T) , (T , T , H) }

Probability of getting one head = 3/8

favorable outcome of getting two head : { (H , H , T) , (H , T , H) , (T , H , H) }

Probability of getting two head = 3/8

Probability of getting exactly one head or two heads = $$3\over 8$$ + $$3\over 8$$ = $$6\over 8$$ = $$3\over 4$$

Hence, Probability of getting exactly one head or two heads = $$3\over 4$$

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