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  • I have a doubt, kindly clarify. An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is :

An unbiased coin is tossed eight times.  The probability of obtaining at least one head and at least one tail is :

 

  • Option 1) (255/256)  
  • Option 2) (127/128) 
  • Option 3) (63/64) 
  • Option 4) (1/2)
 

Answers (1)

best_answer

As we learnt in

Algebra of events -

[ If A and B are mutually exclusive events then A\cap B= \phi

\therefore P\left ( A\cap B \right )= 0

\therefore P\left ( A\cup B \right )= P\left ( A \right )+P\left ( B \right )

\Rightarrow P\left ( A\cup B\cup C \right )= P\left ( A \right )+\left P( B \right )+P\left ( C \right )

\Rightarrow P\left ( \overline{A}\cap \overline{B} \right )= 1-P\left ( A\cup B \right )

\Rightarrow P\left ( \overline{A}\cup \overline{B} \right )= 1-P\left ( A\cap B \right )

\Rightarrow P \left ( A \right )=P\left ( A\cap B \right )+P\left ( A\cap \overline{B} \right )

\Rightarrow P \left ( B\right )=P\left ( B\cap A \right )+P\left ( B\cap \overline{A} \right ) ]

-

 

 Number of throws =8

 

Prob(Atleast one H and atleast one T)

=1-P(All Heads or All Tails)

=1-(All Heads +All Tails)

=1- \left(\frac{1}{256}+\frac{1}{256} \right )

=1-\frac{2}{256}=\frac{1-1}{128}=\frac{127}{128}


Option 1)

\frac{255}{256}

This option is incorrect

Option 2)

\frac{127}{128}

This option is correct

Option 3)

\frac{63}{64}

This option is incorrect

Option 4)

\frac{1}{2}

This option is incorrect

Posted by

divya.saini

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