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Question:
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A coin is tossed three times. Consider the following events:
A: No head appears
B: Exactly one head appears
C: At least two heads appear Which one of the following is correct?
Options:
A: No head appears
B: Exactly one head appears
C: At least two heads appear Which one of the following is correct?
Solution:
2151 Upvotes
Verified Answer
The correct answer is:
$A \cap\left(B^{\prime} \cup C^{\prime}\right)=B^{\prime} \cap C$
$\quad U=\{(H H H)(H H T)(H T H)(H T T)(T H H)(T H T)(T T H)(T T T)\}$
$A=\{(T T T)\}$
$B=\{(H T T)(T H T)(T T H)\}$
$C=\{(H H H)(H H T)(H T H)(T H H)\}$
By checking the options
(d) $A \cap\left(B^{\prime} \cup C^{\prime}\right)=B^{\prime} \cap C^{\prime}$ is correct.
$A=\{(T T T)\}$
$B=\{(H T T)(T H T)(T T H)\}$
$C=\{(H H H)(H H T)(H T H)(T H H)\}$
By checking the options
(d) $A \cap\left(B^{\prime} \cup C^{\prime}\right)=B^{\prime} \cap C^{\prime}$ is correct.
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