Question

Three different numbers are chosen at random without replacement from {1, 2, 3, . 15}, Let E1, be the event that minimum of the chosen numbers is 5 and E2 be the event that their maximum is 10, then​

Answer 1

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Answer 2

The probability that the minimum of the chosen numbers is 5, or their maximum is 10 is $\frac{87}{455}$.

Given:

E1 is the event of getting the minimum number 5 and E2 is the event of getting the minimum number 10.

To Find:

The probability that the minimum of the chosen numbers is 5, or their maximum is 10.

Solution:

p(E1)=p( getting 5 and the other two numbers between 4 to 15)

$\frac{11C_{2} }{15C_{3} } =\frac{11!}{2!9!}*\frac{3!12!}{15!} \\\\= \frac{11*10*9!*3*2!*12!}{2!*9!*15*14*13*12!} \\\\=\frac{11*10*3}{15*14*13}\\\\= \frac{330}{2730} \\\\=\frac{11}{91}$

P(E2)=p( getting 10 and other two numbers between 1 to 9)

$\frac{9C_{2} }{15C_{3} } =\frac{9!}{2!7!}*\frac{3!12!}{15!} \\\\= \frac{9*8*7!*3*2!*12!}{2!*7!*15*14*13*12!} \\\\=\frac{9*8*3}{15*14*13}\\\\= \frac{216}{2730} \\\\=\frac{36}{455}$

P(E1∩E2)p=(getting first number 5, second number 10, and third number between 6 to 9)

$\frac{4C_{1} }{15C_{3} } =\frac{4!}{1!3!}*\frac{3!12!}{15!} \\\\= \frac{4!*3!*12!}{1!*3!*15*14*13*12!} \\\\=\frac{4!}{15*14*13}\\\\= \frac{24}{2730} \\\\=\frac{4}{455}$

P(E1∪E2) = P(E1) +P(E2) - P(E1∩E2)

$=\frac{11}{91} +\frac{36}{455} -\frac{4}{455} \\\\= \frac{11}{91} +\frac{32}{455} \\\\=\frac{55+32}{455} \\\\=\frac{87}{455}$

The probability that the minimum of the chosen numbers is 5, or their maximum is 10 is $\frac{87}{455}$.

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