In a relay race there are five teams $A, \,B, \,C, \,D$ and $E$. What is the probability that $A,\, B$ and $C$ are first three to finish (in any order) (Assume that all finishing orders are equally likely)
If we consider the sample space consisting of all finishing orders in the first three places, we will have $^{5} P _{3},$ i.e., $, \frac{5 \,!}{(5-3) \,!}$ $=5 \times 4 \times 3=60$ sample points, each with a probability of $\frac{1}{60}$.
$A$, $B$ and $C$ are the first three finishers.
There will be $3 \,!$ arrangements for $A, \,B$ and $C$.
Therefore, the sample points corresponding to this event will be $3 \,!$ in number.
So $P( A , \,B $ and $C$ are first three to finish) $=\frac{3\, !}{60}=\frac{6}{60}=\frac{1}{10}$
The probability of hitting a target by three marks men is $\frac{1}{2} , \frac{1}{3}$ and $\frac{1}{4}$ respectively. If the probability that exactly two of them will hit the target is $\lambda$ and that at least two of them hit the target is $\mu$ then $\lambda + \mu$ is equal to :-
Three vertices are chosen randomly from the seven vertices of a regular $7$ -sided polygon. The probability that they form the vertices of an isosceles triangle is
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