3 numbers are chosen at random without replacement from { 1 , 2 , 3 …… .14 } Let Let A = { min of chosen number is 5 } B = { max of chosen no is 11 } P ( A ∪ B ) = 91 K + 11 then K = ____

P ( A ) = 14 C 3 9 C 2 = 91 9 , P ( B ) = 14 C 3 10 C 2 = 364 45 P ( A ∩ B ) = 14 C 3 5 C 1 = 364 5 P ( A ∪ B ) = 91 19

Mathematics - Probability Question with Solution | TestHub

MathematicsProbabilityAddition theoremHard2 minQB

MathematicsHardinteger

3 numbers are chosen at random without replacement from ${1, 2, 3 \dots\dots .14}$ Let Let $A = {min$ of chosen number is 5 $}$ $B = {$ max of chosen no is 11 $}$

$P (A \cup B) = \frac{K + 11}{91}$ then $K =$ $____$

Answer:

Solution:

$P (A) = \frac{^{9} C _{2}}{^{14} C _{3}} = \frac{9}{91}, P (B) = \frac{^{10} C _{2}}{^{14} C _{3}} = \frac{45}{364}$

$P (A \cap B) = \frac{^{5} C _{1}}{^{14} C _{3}} = \frac{5}{364} P (A \cup B) = \frac{19}{91}$

Stream:JEESubject:MathematicsTopic:ProbabilitySubtopic:Addition theorem

⏱ 2mℹ️ Source: QB

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