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  • A,B,C try to hit a target simultaneously but independently. their respective probablities of hitting the targets are <img alt="\frac{3}{4},\frac{1}{2},\frac{5}{8}." src="http://entrancecorner.

A,B,C try to hit a target simultaneously but independently. their respective probablities of hitting the targets are \frac{3}{4},\frac{1}{2},\frac{5}{8}. The probablity that the target is hit by A or B but only by C is 

Option: 1

\frac{21}{64}


Option: 2

\frac{7}{8}


Option: 3

\frac{7}{32}


Option: 4

\frac{9}{64}


Answers (1)

best_answer

\\\mathrm{P}(\mathrm{A} \text { or } \mathrm{B} \text { but not } \mathrm{by}\; \mathrm{C})=\mathrm{P}((\mathrm{A} \cup \mathrm{B}) \cap \overline{\mathrm{C}}) \\ =\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \times \mathrm{P}(\overline{\mathrm{C}}) \\ =[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})] \times \mathrm{P}(\overline{\mathrm{C}}) \\ \\= {\left[\frac{3}{4}+\frac{1}{2}-\frac{3}{4} \times \frac{1}{2}\right] \times \frac{3}{8}=\left(\frac{6+4-3}{8}\right) \times \frac{3}{8}=\frac{21}{64}}

Posted by

Pankaj Sanodiya

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