A box contains 9 tickets numbered 1 to 9 inclusive. If 3 tickets are drawn from the box one at a time, the probability that they are alternatively either odd, even, odd or even, odd, even is

Question

A box contains 9 tickets numbered 1 to 9 inclusive. If 3 tickets are drawn from the box one at a time, the probability that they are alternatively either odd, even, odd or even, odd, even is (A) (5/17) (B) (4/17) (C) (5/16) (D) (5/18)

Tardigrade Admin · Accepted Answer

No. of tickets = 9 No. of odd numbered tickets = 5 No. of even numbered tickets = 4 Required probability = P odd, even, odd + P (even, odd, even) =(5C1/9C1)×(4C1/8C1)×(4C1/7C1)×(4C1/9C1)×(5C1/8C1)×(3C1/7C1) =(5/9)×(4/8)×(4/7)× (4/9)× (5/8)× (3/7)= (5/18)

Q. A box contains $9$ tickets numbered $1$ to $9$ inclusive. If $3$ tickets are drawn from the box one at a time, the probability that they are alternatively either {odd, even, odd} or {even, odd, even} is

$\frac{5}{17}$

$\frac{4}{17}$

$\frac{5}{16}$

$\frac{5}{18}$

Q. A box contains 9 tickets numbered 1 to 9 inclusive. If 3 tickets are drawn from the box one at a time, the probability that they are alternatively either {odd, even, odd} or {even, odd, even} is

175​

174​

165​

185​

No. of tickets = 9 No. of odd numbered tickets = 5 No. of even numbered tickets = 4 Required probability = P {odd, even, odd} + P (even, odd, even) =9C1​5C1​​×8C1​4C1​​×7C1​4C1​​×9C1​4C1​​×8C1​5C1​​×7C1​3C1​​ =95​×84​×74​×94​×85​×73​=185​

Q. A box contains $9$ tickets numbered $1$ to $9$ inclusive. If $3$ tickets are drawn from the box one at a time, the probability that they are alternatively either {odd, even, odd} or {even, odd, even} is

$\frac{5}{17}$

$\frac{4}{17}$

$\frac{5}{16}$

$\frac{5}{18}$