related to monkey problem ie just like there is 1million penguin .
in that 0.63 saw some sharks and 0.65 saw frogs then what is the probability of seeing both frogs and sharks ???

like this but in exam there are no sharks and frogs ..i just taken some examples

• in place of o.63 it was 0.5....
  and sine both are independent therfore ans will be 0.5*.65 ie 32.5 and it was asked minimum so ans wd be 30...
• 9 years agoHelpfull: Yes(5) No(1)
• 33%
  is the answer
  
• 9 years agoHelpfull: Yes(4) No(1)
• p(shark)=0.63 and p(frogs)=0.65
  both are independent events .
  so, p(shark+frogs)=p(shark)*p(frogs)=0.63*0.65
  this will be the answer.
• 9 years agoHelpfull: Yes(3) No(1)
• min 15%
  let take the total no. is 100
  p(shark)=50, P(frogs)=65
  100-(50+65)=15
  so 15 will be the min %.
• 9 years agoHelpfull: Yes(1) No(0)
• @NITIN can u xplain...question was asking minimum...
• 9 years agoHelpfull: Yes(0) No(0)
• ya 30 is the right one
  @rozzer coz we need to take minimum
• 9 years agoHelpfull: Yes(0) No(1)
• the nos in the question was 0.5 nd 0.65 nd qstn was what is the minimum probability of seeing both??
  option:
  a> 0%
  b>15%
  c> & d> nt remember
  
• 9 years agoHelpfull: Yes(0) No(0)
• intersection of both will be the minimum probability
• 9 years agoHelpfull: Yes(0) No(0)