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  • friends. They decide to meet between 1 <img alt="\mathrm{\mat

\mathrm{A \: and \: B} friends. They decide to meet between 1 \mathrm{\mathrm{PM} \, \: and \: \: 2 \mathrm{PM}} on a given day. There is a condition that whoever arrives first will not wait for the other for more than 15 minutes. The probability that they will meet on that day is
 

Option: 1

\frac{1}{4}


Option: 2

\frac{1}{16}


Option: 3

\frac{7}{16}


Option: 4

\frac{9}{16}


Answers (1)

best_answer

Method I:
Case I: If A come earlier than B, then P (He will not meet with B)
=\frac{45}{60}
Case II : If B come earlier than A, then P ( He will not meel with A)
=\frac{45}{60}

So, P (when they will not meet)

=\frac{45}{60} \times \frac{45}{60}=\frac{9}{16}

So, P (when they meet between 1 and 2 p.m.) = 1 - P (they will not meet)
=1-\frac{9}{16}=\frac{7}{16}

The shaded graph shows the chances that A and B will meet

\therefore The probability that they will meet

\mathrm{ =\frac{\text { Area AOBCDEA }}{\text { AreaOXDYO }} }

\mathrm{ =\frac{1-\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4} \times 2}{1} }

\mathrm{=1-\frac{9}{16}=\frac{7}{16}}




 

 

Posted by

Riya

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