Coin is twiced twice.What is probability of getting 2 consecutive tails
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Hi there!
Here's the answer :
Let S be sample space
n(S) = Total No. of outcomes when coin is tossed twice
n(S) = 2² = 4
[ °•° S= {(H,T) , (T,H) , (H,H) & (T,T} ]
E be the event of getting Consecutive Tails
n(E) = No. of ways possible for getting consecutive tails.
n(E)= 1
[ °•° Only one possible case ( i.e., (T,T) ) in set S satisfying Event 'E' such that there are two tails consecutively. ]
Probability of Occurrence of Event
= (No. of Possible outcomes) / (Total No. of Outcomes)
= n(E) / n(S)
•°• Required probability= 1/4
:)
Hope it helps
Here's the answer :
Let S be sample space
n(S) = Total No. of outcomes when coin is tossed twice
n(S) = 2² = 4
[ °•° S= {(H,T) , (T,H) , (H,H) & (T,T} ]
E be the event of getting Consecutive Tails
n(E) = No. of ways possible for getting consecutive tails.
n(E)= 1
[ °•° Only one possible case ( i.e., (T,T) ) in set S satisfying Event 'E' such that there are two tails consecutively. ]
Probability of Occurrence of Event
= (No. of Possible outcomes) / (Total No. of Outcomes)
= n(E) / n(S)
•°• Required probability= 1/4
:)
Hope it helps
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