A person forgets two digits of user ID for a website. He remembers that two digits are odd. What is the probability of him typing the correct last digits by randomly typing 2 odd digits?
(1/25) (1/5) (1/2) (2/5)

• 1/25
  As only one correct from total 5*5 cases
• 7 years agoHelpfull: Yes(15) No(1)
• The odd digits are odd or. 1,3,5,7,9
  And two digits can be same ,so probability is (1/5)*(1/5)=1/25
• 7 years agoHelpfull: Yes(10) No(1)
• No of odd digits between 0 to 9 is= 5 (1,3,5,7,9)
  Favorable case of selecting 2 odd digits is p(A) = 5C2 = 10
  
  and sample case of arranging two odd digits is p(S) = 5*4 = 20
  
  So,
  Probability p(N) = p(A) / p(S) = 10 / 20 = 1/2 Answer
• 7 years agoHelpfull: Yes(3) No(15)
• no of odd digits =1,3,5,7,9
  sample space 5*5 as they said only odd the number it might be repeated.
  p=5c2/5*5=2/5.
• 7 years agoHelpfull: Yes(1) No(7)
• probability= 5c2/25= 2/5
• 4 years agoHelpfull: Yes(1) No(0)
• The total possibilty of selecting 2 numbers in a way(11,13,15,....) = 5*5
  Now, Out of these only 1 is correct.
  Prob = Favourable Outcome/ Total Outcome
  Ans = 1/25
• 3 years agoHelpfull: Yes(0) No(0)
• No of odd digits between 0 to 9 is= 5 (1,3,5,7,9)
  Favorable case of selecting 2 odd digits is p(A) = 5C2 = 10
  
  and sample case of arranging two odd digits is p(S) = 5*4 = 20
  So, Probability p(N) = p(A) / p(S) = 10 / 20 = 1/2 Answer
• 6 Months agoHelpfull: Yes(0) No(0)