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If three numbers are drawn at random successively without replacement from a set $S=\{1,2, \ldots 10\}$, then the probability that the minimum of the chosen numbers is 3 or their maximum is 7 .
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Verified Answer
The correct answer is:
$\frac{11}{40}$
Given, set $S=\{1,2, \ldots \ldots, 10\}$
Here, three numbers are drawn at random from the given set. So, total possible outcomes, $n={ }^{10} C_3=120$
Let $\mathrm{A}$ be the event that minimum of chosen number is 3 .
$$
\therefore \quad n(A)=\{4,5,6,7,8,9,10\}={ }^7 C_2=21
$$
$\mathrm{B}$ be the event that maximum at chosen number is 7 .
$$
\therefore \quad n(B)=\{1,2,3,4,5,6\}={ }^6 C_2=15
$$
So, $n(A \cap B)=\{4,5,6\}={ }^3 C_1=3$
Hence, required probability
$$
\begin{aligned}
& P=\frac{n(A)+n(B)-n(A \cap B)}{n}=\frac{21+15-3}{120} \\
& P=\frac{11}{40}
\end{aligned}
$$
Here, three numbers are drawn at random from the given set. So, total possible outcomes, $n={ }^{10} C_3=120$
Let $\mathrm{A}$ be the event that minimum of chosen number is 3 .
$$
\therefore \quad n(A)=\{4,5,6,7,8,9,10\}={ }^7 C_2=21
$$
$\mathrm{B}$ be the event that maximum at chosen number is 7 .
$$
\therefore \quad n(B)=\{1,2,3,4,5,6\}={ }^6 C_2=15
$$
So, $n(A \cap B)=\{4,5,6\}={ }^3 C_1=3$
Hence, required probability
$$
\begin{aligned}
& P=\frac{n(A)+n(B)-n(A \cap B)}{n}=\frac{21+15-3}{120} \\
& P=\frac{11}{40}
\end{aligned}
$$
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