A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{8}\]
D) \[\frac{3}{4}\]
Correct Answer: D
Solution :
Total number of ways \[=(HH,\,HT,\,TH,\,TT)\] \[P\] (head on first toss) \[=\frac{2}{4}=\frac{1}{2}=P(A)\] \[P\] (head on second toss) \[=\frac{2}{4}=\frac{1}{2}=P(B)\] \[P\] (head on both toss) \[=\frac{1}{4}=P(A\cap B)\] Hence required probability is, \[P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac{1}{2}+\frac{1}{2}-\frac{1}{4}=\frac{3}{4}\] .
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