Three number are chosen at random without replacement from {1, 2, 3,...10}. The probability that minimum of the chosen number is 3 or their maximum is 7, cannot exceed

Engineering

Mathematics

Conditional Probability

Question

11/30

11/40

1/7

1/8

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Let A and B denote the following events:

A: minimum of the chosen number is 3.

B: maximum of the chose number is 7.

We have P(A) = P (choosing 3 and two other numbers from 4 to 10)

$= \frac{^{7} C_{2}}{^{10} C_{3}} = \frac{7 \times 6}{2} \times \frac{3 \times 2}{10 \times 9 \times 8} = \frac{7}{40}$

$P(B) = P$ (choosing 7 and two other numbers from 1 to 6)

$= \frac{^{6} C_{2}}{^{10} C_{3}} = \frac{6 \times 5}{2} \times \frac{3 \times 2}{10 \times 9 \times 8} = \frac{1}{8}$

$P(A⋂B) = P$ (choosing 3 and 7 and one other number from 4 to 6).

$= \frac{3}{^{10} C_{3}} = \frac{3 \times 3 \times 2}{10 \times 9 \times 8} = \frac{1}{40}$ .

Now, P(AUB) = P(A) + P(B) – P(A⋂B)

$= \frac{7}{40} + \frac{1}{8} - \frac{1}{40}$ $= \frac{11}{40}$ , $\frac{11}{30}$

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