Three persons PQ and R independently try to hit a target.  If the probabilities of their hitting the target are \small \frac{3}{4} ,\frac{1}{2} and\frac{5}{8}   respectively, then the probability that the target is hit by P or Q but not by R is :

  • Option 1)

    \frac{21}{64}

  • Option 2)

    \frac{9}{64}

  • Option 3)

    \frac{15}{64}

  • Option 4)

    \frac{39}{64}    

 

Answers (1)

As we learnt in 

Addition Theorem of Probability -

P\left ( A\cup B \right )= P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )

in general:

P\left ( A_{1}\cup A_{2}\cup A_{3}\cdots A_{n} \right )=\sum_{i=1 }^{n}P\left ( A_{i} \right )-\sum_{i< j}^{n}P\left ( A_{i}\cap A_{j} \right )+\sum_{i< j< k}^{n} P\left ( A_{i}\cap A_{j}\cap A_{k} \right )-\cdots -\left ( -1 \right )^{n-1}P\left ( A_{1}\cap A_{2}\cap A_{3}\cdots \cap A_{n} \right )

-

 

 

Independent events -

Two or more events are said to be independent if occurence or non occurence of any of them does not affect the probability of occurence of or non - occurence of other events.

-

 

 

P(P)=\frac{3}{4};P(Q)=\frac{1}{2};P(R)=\frac{5}{8}

P(P\ or\ Q)=P(P)+P(Q)-P(P\cap Q)

=\frac{3}{4}+\frac{1}{2}-\frac{3}{8}

=\frac{7}{8}

P(\bar{R})=\frac{3}{4}

P (P or Q not R) =\frac{7}{8}\times \frac{3}{8}=\frac{21}{16}


Option 1)

\frac{21}{64}

This is correct

Option 2)

\frac{9}{64}

This is incorrect

Option 3)

\frac{15}{64}

This is incorrect

Option 4)

\frac{39}{64}    

This is incorrect

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