Statistics And Probability Question 312

Question: A coin is tossed twice. If events A and B are defined as : A = head on first toss, $ B= $ head on second toss. Then the probability of $ A\cup B= $

Options:

A) $ \frac{1}{4} $

B) $ \frac{1}{2} $

C) $ \frac{1}{8} $

D) $ \frac{3}{4} $

Show Answer

Answer:

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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