@chandrakant.k@Brooklynmera logic...3 numbers me se agar 1 number ka bhi unit digit 5 ya 0 hui,,toh it ll be divisible by 5..like 13 x 14 x 15 or 13 x 14 x 20 ...so 3C2 * 2/10 * 10/10*10/10 = 0.6@chandrakant.k only 1 number ka ho toh hi chalega na..all 3 jaruri thdi he..they can be
flaw in ur logic: u dont know how much digit the number is , ie number can 2 digits, 3 digit so on which would change ur ans
: no your missing the point. Lets say 3 nos are :ab5, def, kl u get a value now lets tak no as: x0, d, a ur ans changes
bhai ans kese change hga...samaj nhi aa rha
probaility of numbers 0-9 to come at unit digit is equal...we want any one out of 3 numbers to have either 0 or 5 at unit place...so ab5, def, kl and x0 , d, a me as u qouted ,unit digit kch bhi ho na y r we intrsted in it..dimag kharab ho chuka he
Let an = 1111111.....1, where 1 occurs n number of time. Then, i. a741 is not a prime. ii a534 is not a prime. iii a123 is not a prime. iv a77 is not a prime. A. (i) is correct. B. (i) and (ii) are correct. C. (ii) and (iii) are correct. D. All of them are correct. E. None of them is correct.
Consider the total number of numbers to be 5NNow, each number can be represented as 5k, 5k+ 1, 5k + 2, 5k +3, 5k + 4So, every form (i.e. 5k + 1 and so on) has N numbers in it So, probability that the product of three numbers is not divisible by 5 = (4N/5N)^3 = 0.512So, Probability required = 1 - 0.512 = 0.488 ?
answer seems to be right as shown by @chillfactor sir but
How many minimum randomly arranged boys do we need to have such that if anyone of them is selected then there is at least 60% chance that he was born in a leap year.? (a) 3 (b) 4 (c) 5 (d) never possible.
