A) 0
B) 1
C) \[\frac{1}{2}\]
D) \[\frac{1}{4}\]
Correct Answer: C
Solution :
Let E be the event of getting at least two heads. |
\[\therefore \,\,\,\,\,\,\,\,E=\{HHT,HTH,THH,HHH\}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n(E)=4\] |
\[\therefore \] Required probability \[=\frac{n(E)}{n(S)}=\frac{4}{8}=\frac{1}{2}\] |
So, Option [c] is correct. |
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