A bag contains 17 tickets numbered from 1 to 17. A ticket is draw at random, then another ticket is drawn at random, then another ticket is drawn without replacing the first one . The probability that both the tickets may show even numbers is

Question

A bag contains 17 tickets numbered from 1 to 17. A ticket is draw at random, then another ticket is drawn at random, then another ticket is drawn without replacing the first one . The probability that both the tickets may show even numbers is (A) (7/34) (B) (8/17) (C) (7/16) (D) (7/17)

Tardigrade Admin · Accepted Answer

Required probability = (8/17) × (7/16) = (7/34)

Q. A bag contains 17 tickets numbered from 1 to 17. A ticket is draw at random, then another ticket is drawn at random, then another ticket is drawn without replacing the first one . The probability that both the tickets may show even numbers is

$\frac{7}{34}$

$\frac{8}{17}$

$\frac{7}{16}$

$\frac{7}{17}$

Required probability = $\frac{8}{17} \times \frac{7}{16} = \frac{7}{34}$

Q. A bag contains 17 tickets numbered from 1 to 17. A ticket is draw at random, then another ticket is drawn at random, then another ticket is drawn without replacing the first one . The probability that both the tickets may show even numbers is

347​

178​

167​

177​

Required probability = 178​×167​=347​

$\frac{7}{34}$

$\frac{8}{17}$

$\frac{7}{16}$

$\frac{7}{17}$

Required probability = $\frac{8}{17} \times \frac{7}{16} = \frac{7}{34}$