Probability Question 245
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} $ .
BETA
