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

Let A be the event that maximum is 6 . B be the event that minimum is 3 . P A = C 2 5 C 3 8 (number less than 6 is 5 ) P B = C 2 5 C 3 8 (number greater than 3 is 5 ) P A ∩ B = C 1 2 C 3 8 ∴ Required probability = P B / A = P A ∩ B P A = C 1 2 C 2 5 = 2 10 = 1 5

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MathematicsProbabilityGeneral (Event SS, Definitions, Odd against/Favour)Easy2 minPYQ_2019

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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

Options:

Answer:

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

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

Stream:BITSATSubject:MathematicsTopic:ProbabilitySubtopic:General (Event SS, Definitions, Odd against/Favour)

⏱ 2mℹ️ Source: PYQ_2019

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