Question 369672
Look at all of the possible outcomes for 3 excursions-{{{C}}} is catches, {{{D}}} is doesn't catch.
There are {{{2^3}}} possible outcomes.
{{{CCC}}}
{{{CCD}}}
{{{CDC}}}
{{{CDD}}}
{{{DCC}}}
{{{DCD}}}
{{{DDC}}}
{{{DDD}}}
Only one outcomes has him not catching at least one, DDD
{{{P(D)=1-0.15=0.85}}}
{{{P(DDD)=(0.85)(0.85)(0.85)=0.614125}}}
P(DDD)+P(at least 1 C)={{{1}}}
P(at least 1 C)={{{1-P(DDD)}}}
P(at least 1 C)={{{1-0.614125}}}
P(at least 1 C)={{{highlight(0.385875)}}}