Respuesta :
Sample Space ={5 Red, 8White, 2 Blacks} = 15 possible outcomes.
Favorable outcomes={5 Red}
1) P(drawing 1st Red) = 5/15
2) P(drawing ANOTHER RED) = 4/14 (because one Red already drawn and NOT replaced, so the remaining Red are now 4 and the new sample space is now 14, since 1 red already drawn), Hence
P( 1 Red AND another Red) = (5/15).(4/15) or:
"5 over 15 multiplied by 4 over 14"
Favorable outcomes={5 Red}
1) P(drawing 1st Red) = 5/15
2) P(drawing ANOTHER RED) = 4/14 (because one Red already drawn and NOT replaced, so the remaining Red are now 4 and the new sample space is now 14, since 1 red already drawn), Hence
P( 1 Red AND another Red) = (5/15).(4/15) or:
"5 over 15 multiplied by 4 over 14"