For a party 8 guests are invited by a husband and his wife. They sit in a row for dinner. The probability that the husband and his wife sit together is

Question

For a party 8 guests are invited by a husband and his wife. They sit in a row for dinner. The probability that the husband and his wife sit together is (A)  (2/7)  (B)  (2/9)  (C)  (1/9)  (D)  (4/9)

Tardigrade Admin · Accepted Answer

Total number of way of sitting =9 ! and number of favourable ways of sitting =2 × 8 ! ∴ Required probability =(2 × 8 !/9 !)=(2/9).

Q. For a party $8$ guests are invited by a husband and his wife. They sit in a row for dinner. The probability that the husband and his wife sit together is

$\frac{2}{7}$

$\frac{2}{9}$

$\frac{1}{9}$

$\frac{4}{9}$

Total number of way of sitting $= 9!$
and number of favourable ways of sitting $= 2 \times 8!$
$∴$ Required probability $= \frac{2 \times 8 !}{9 !} = \frac{2}{9}$ .

Q. For a party 8 guests are invited by a husband and his wife. They sit in a row for dinner. The probability that the husband and his wife sit together is

72​

92​

91​

94​