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14.Probability
hard
Fifteen persons among whom are $A$ and $B$, sit down at random at a round table. The probability that there are $4$ persons between $A$ and $B$, is
A
$\frac{1}{3}$
B
$\frac{2}{3}$
C
$\frac{2}{7}$
D
$\frac{1}{7}$
Solution
(d) Let $A$ occupy any seat at the round table. Then there are $14$ seats available for $B$.
If there are to be four persons between $A$ and $B.$
Then $B$ has only two ways to sit, as show in the fig.
Hence required probability $ = \frac{2}{{14}} = \frac{1}{7}.$
Standard 11
Mathematics
