Probability Ques 31
Three numbers are chosen at random without replacement from $\{1,2, \ldots, 10\}$. The probability that the minimum of the chosen number is $3$ , or their maximum is $7$ , is … .
(1997C, 2M)
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Answer:
Correct Answer: 31.$(\frac{11}{40})$
Solution:
- Let $E _1$ be the event getting minimum number $3$ and $E _2$ be the event getting maximum number $ 7$ .
Then, $P\left(E _1\right)=P$ (getting one number $3$ and other two from numbers $4$ to $10$)
$ =\frac{{ }^{1} C _1 \times{ }^{7} C _2}{{ }^{10} C _3}=\frac{7}{40} $
$P\left(E _2\right)=P$ (getting one number $7$ and other two from numbers $1$ to $6$ )
$ =\frac{{ }^{1} C _1 \times{ }^{6} C _2}{{ }^{10} C _3}=\frac{1}{8} $
and $P\left(E _1 \cap E _2\right)=P$ ( getting one number $3$ , second number $7$ and third from $4$ to $6$ )
$ \begin{aligned} & =\frac{{ }^{1} C _1 \times{ }^{1} C _1 \times{ }^{3} C _1}{{ }^{10} C _3}=\frac{1}{40} \\ \therefore \quad P\left(E _1 \cup E _2\right) & =P\left(E _1\right)+P\left(E _2\right)-P\left(E _1 \cap E _2\right) \\ & =\frac{7}{40}+\frac{1}{8}-\frac{1}{40}=\frac{11}{40} \end{aligned} $
BETA
