Three numbers are chosen at random without replacement from {1,2,3,……,8}. The probability that their minimum is 3, given that their maximum is 6, is

Correct answer: (B) 1/5. Let A be the event that maximum is 6. B be the event that minimum is 3. PA=C25/C38 (number less than 6 is 5) PB=C25/C38 (number greater than 3 is 5) PA∩B=C12/C38 ∴ Required probability =PB/A=(PA∩B)/PA=C12/C25=2/10=1/5

Q60 / 125 · BITSAT 2019 · Mathematics

Three numbers are chosen at random without replacement from ${1, 2, 3, \dots \dots, 8}$ . The probability that their minimum is $3$ , given that their maximum is $6$ , is

A
$\frac{3}{8}$
Correct answer
B
$\frac{1}{5}$
C
$\frac{1}{4}$
D
$\frac{2}{5}$

Solution

Let $A$ be the event that maximum is $6$ .

$B$ be the event that minimum is $3$ .

$P (A) = \frac{{}^{5}C_{2}}{{}^{8}C_{3}}$ (number less than $6$ is $5$ )

$P (B) = \frac{{}^{5}C_{2}}{{}^{8}C_{3}}$ (number greater than $3$ is $5$ )

$P (A \cap B) = \frac{{}^{2}C_{1}}{{}^{8}C_{3}}$

$∴$ Required probability

$= P (B / A) = \frac{P (A \cap B)}{P (A)} = \frac{{}^{2}C_{1}}{{}^{5}C_{2}} = \frac{2}{10} = \frac{1}{5}$

Hence, the correct answer is option: 'B'.

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