100 students appeared for two different examinations 60 passed the first,50 the second and 30 both the examinations.Find the probability that a student selected at random failed in both the examination
a)5/6 b)1/5 c)1/7 d)5/7

• 60+50-30=80
  100-80=20
  20/100=1/5.
  so B is the answer.
• 8 years agoHelpfull: Yes(27) No(1)
• 100 students
  60 pass(1)&50 pass(2)
  30 pass (both 1&2)
  p(aub)=60+50-30=80
  probability of selecting a passed student =80/100
  not selecting=(1-80/100)=20/100=1/5
  
  
• 8 years agoHelpfull: Yes(9) No(0)
• b ,in first round the probability of failures is 40/100 in second round it is 50/100. the product of these two gives 1/5 as answer
• 8 years agoHelpfull: Yes(5) No(1)
• n(1)=60[no.of students passed in first exam]
  n(2)=50[no.of students passed in second exam]
  n(1∩2)=30[no.of students passed in both exams]
  [n(AUB)=n(A)+n(B)-n(A∩B)]
  no. of students passed = (60+50)-30
  =80
  no. of students failed in both exams = 100 - 80
  =20
  PROBABILITY = 20/100
  =1/5
• 8 years agoHelpfull: Yes(5) No(0)
• first subject 60 pass and 40 fail
  second subject 50 pass and 50 fail
  bot pass 30
  first only pass 60-30=30
  second only pass 50-30=20
  total pass 30+30+20=80
  both fail=100-80=20
  so 20/100=1/5
• 6 years agoHelpfull: Yes(5) No(0)
• We are needed to find (A' intersection B')= (1-60/100)(1-50/100)
  1/5 ans
• 8 years agoHelpfull: Yes(2) No(0)
• Answ is 1/5
• 8 years agoHelpfull: Yes(1) No(0)
• 20c1/110c1=1/5
• 8 years agoHelpfull: Yes(1) No(1)
• 20c1/100c1=1/5
• 8 years agoHelpfull: Yes(1) No(1)
• total students are 100
  30 pased first 20 passed second 30 passed both by venn diagram
  30+30+20=80
  100-80=20
  20/100=1/5
  so, ans is 5
• 7 years agoHelpfull: Yes(0) No(0)
• 60+50-30=80
  100-80=20
  20/100=1/5
  ans: (b) 1/5
• 4 years agoHelpfull: Yes(0) No(0)
• p(aub)=p(a)+a(b)-p(a-b)
  p(aub)=0.6+0.5-0.3
  p(aub)=0.8
  p(aub)'=1-p(aub)
  p(aub)'=1-0.8
  p(aub)'=0.2
• 2 years agoHelpfull: Yes(0) No(0)