Probability 2 Question 10
10. 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: 10. $13.9 %$
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
$$ =\frac{{ }^{1} C _1 \times{ }^{7} C _2}{{ }^{10} C _3}=\frac{7}{40} $$
from numbers 4 to 10)
$P\left(E _2\right)=P$ (getting one number 7 and other two from
$$ =\frac{{ }^{1} C _1 \times{ }^{6} C _2}{{ }^{10} C _3}=\frac{1}{8} $$
numbers 1 to 6 )
and $P\left(E _1 \cap E _2\right)=P($ getting one number 3 , second number 7 and third from 4 to 6 )
$$ \begin{aligned} & =\frac{{ }^{1} C _1 \times{ }^{1} C _1 \times{ }^{3} C _1}{{ }^{10} C _3}=\frac{1}{40} \\ \therefore \quad P\left(E _1 \cup E _2\right) & =P\left(E _1\right)+P\left(E _2\right)-P\left(E _1 \cap E _2\right) \\ & =\frac{7}{40}+\frac{1}{8}-\frac{1}{40}=\frac{11}{40} \end{aligned} $$
