A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{3}\]
D) \[\frac{1}{8}\]
Correct Answer: B
Solution :
(b): Here \[S=\{TTT,TTH,THT,HTT,THH,HTH,HHT,HHH\}\] Let E = event of getting at least two heads = x\[\{THH,HTH,HHT,HHH\}.\] \[\therefore p(E)=\frac{n(E)}{n(S)}=\frac{4}{8}=\frac{1}{2}\]You need to login to perform this action.
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